
Class T u/ /^ 

Book 1E±l 

Copyright N° 



COPYRIGHT DEPOSIT. 



INDUSTRIAL MATHEMATICS 
PRACTICALLY APPLIED 



AN INSTRUCTION AND REFERENCE 
BOOK FOR STUDENTS IN MANUAL 
TRAINING, INDUSTRIAL AND TECHNI- 
CAL SCHOOLS, AND FOR HOME 
STUDY 

BY 

PAUL V. FARNSWORTH 

H 

FORMERLY SUPERVISOR OF THE CADILLAC SCHOOL OF APPLIED MECHANICS 



WITH 250 ILLUSTRATIONS AND 
OVER 1000 PROBLEMS AND ANSWERS 




NEW YORK 

D. VAN NOSTRAND COMPANY 

Eight Warren Street 

1921 

4 2 <$y 



,?4 



Copyright, 1921 
By D. Van Nostrand Company 



OCT 25 1921 



PRINTED IN THE UNITED STATES OP AMERICA 



©CI.A627414 



PREFACE 

This book is the result of 12 years experience as supervisor 
of apprentices, designing, shop foreman and instructor in 
evening classes in technical schools. 

The author has made every effort to simplify and analyze, 
step by step, the various phases of industrial mathematics. 
The problems, examples and illustrations have been carefully 
selected and chosen to make them as practical and interesting 
as possible so that they will stimulate and encourage the 
student to think clearly, reason and analyze for himself. 

The first few pages are intended to be used for review or 
reference work at the discretion of the student, or for those 
who have only an elementary knowledge of mathematics. 

Standard formulas and data, such as are usually found in 
mechanics' hand books, trade journals and technical litera- 
ture, are frequently used and sufficient material in the way 
of explanations, illustrations and examples are given along 
with each subject so that the student will become familiar 
with them. It is hoped that such information will make this 
book of permanent value for reference and as a hand book. 

Paul V. Farnsworth. 
Detroit, Mich. 
May 5, 1921. 



CONTEXTS 
Part I 

Page 

Signs, symbols, abbreviations, etc I 

Notation and numeration 2 

Addition, subtraction, multiplication and division 4 

Cancellation and least common multiples . 9 

Common fractions 10 

Addition, subtraction, multiplication and division of 

fractions 11 

Decimal fractions 15 

Addition, subtraction, multiplication and division of 

decimals 16 

Percentage 19 

Weights and measures 20 

Ratio and proportion 24 

Taper calculation 26 

Interest 28 

Pulley and gear diameters 30 

Square root, involution and evolution 34 

Cube root 37 

The circle 39 

Mensuration and geometry 43 

Review exercises 51 

Part II 

Formulas and algebraical expressions 52 

Progression 60 

Trigonometry 64 

Trigonometrical functions 67 

Feeds and speeds 74 

v 



VI CONTENTS 

Cost calculation jj 

Levers 88 

Pulleys t)i 

Screws 94 

Inclined planes 95 

.Wedges 97 

Gearing definitions, etc 98 

Spur gearing 101 

Bevel gearing 1 05 

Worm gearing 108 

Spiral gearing in 

Review exercises 114 

Part III 

Dovetail slides 116 

Screw threads 117 

Lathe change gears 121 

Indexing (simple, compound, differential and angular) . . 124 

Spiral milling 129 

Friction 130 

Electricity 133 

Work, power and the steam engine 135 

Strength and proportions of gear teeth 138 

Resolution of forces 140 

Falling bodies. 142 

Centrifugal force 145 

Horse power of belting 146 

Length of belting 147 

Rope drives . 149 

Cable or wire rope drives 150 

Chain transmission 152 

Shaft design 154 

Bearing design 156 

Ball Bearing Design 159 



CONTENTS Vll 

Center of gravity, radius of gyration and moment of 
inertia 161 

Part IV 

Graphical charts 166 

Strength of materials 175 

Springs 187 

Pipes and cylinders 191 

Riveted joints 194 

Logarithms 196 

Heat 200 

Metal cutting , 205 

Force, work, energy and momentum 210 

Force, shear and bending moment diagrams 216 

Pendulum 226 

Cam design 231 

Review exerc ses 236 

Appendix 

Decimal equivalents, squares and roots of fractions. 

(Table I) 238 

Natural trigonometrical functions (Table II) 239 

Common logarithms (Table III) 242 

Specific gravity of materials (Table IV) 243 

Weight and specific gravity of liquids (Table V) 244 

Melting point of materials (Table VI) 244 

Strength of miscellaneous metals . (Table VII) 245 

Tapers and angles (Table VIII) 245 

Cutting speeds (Table IX) 246 

Weights and areas of round, square and hexagon steel. 

(Table X) 248 

Circumference and area and circles (Table XI) 251 

Standard dimensions of wrought iron welded tubes. 

(Table XII) 251 



Vlll CONTENTS 

Tap drill sizes (Table XIII) 252 

Twist drill and steel wire gages (Table XIV) 253 

Standard key seats (Table XV) 254 

Multiplication tables (Table XVI) 255 

Answers to exercises 257 




Drilling Machine 
A — Feed box. F — Column. 

B — Back-gear lever. G — Cone pulley. 

C — Hand feed wheel. H — Back-gears. 

D — Spindle. I — Reverse lever. 

E— Work-table. J— Feed trip. 




Universal Gear Cutter 




Plain Milling Machine 



PART I 

Signs, Symbols and Abbreviations Used in Mathematics in 
General, with Illustrations of Their Uses 

= signifies equals, 7 and 2=9. 

+ = plus, 7 + 2=9 

— = minus, 9 — 7=2. 

X = times, 3X2 = 6. 

-5- = divided by, 64-2=3. Note the alternative 

usage, 6/2 = 3. 
.'. = therefore, if 2+2 =4.'. 4— 2 = 2. 

: = is to, 2 : 4 as 3 : 6. 
: : = as, 2 : 4 : : 3 : 6. 
> = greater than, 7 > 3. 
< = less than 3 < 7. 
V or yj = radical sign or square root, V4 = 2, or V4 = 2. 
yl = cube root, V8 = 2. 
yj = fifth root, V32 = 2. 

4 2 = four squared (second power of 4) 4 2 = 4 X 4 

= 16. 

4 3 = four cubed (3d power of 4) 4 3 = 4 X 4 X 4 = 64. 
log = logarithm of, log 2056 = 3.3131. 

7r = pi or 3.1416 or (3.14159265359) or approximately 

3|, 2 X 7T = 6.2832. 
g = acceleration of gravity (32.16 ft. per sec. per 

sec), 2 X g = 64.32. 
tan = tangent of, tan 20 = 0.36397. 
cot = cotangent of, cot 20 = 2.7475. 
sin = sine of, sin 20 = 0.34202. 
cos = cosine of, cos 20 = 0.93969. 
— = vinculum, 2X4 + 2 = 2X6 = 12. 
( ) = parentheses, 2X (3 + 2) -7-5 = 2X5^5 = 2. 



.2 INDUSTRIAL MATHEMATICS 

{ } braces, = 24 -4- {2 X (4 + 2)} = 2. 

[ ] = brackets, [3 X {2 X (4 + 2)}] X 2 = 72. 

These signs (vinculum, parentheses, braces and 
brackets) denote that the quantities included 
within them are to be treated as a whole. 
/ = angle, The Z of the U.S.S. thread is 6o°. 
L = right angle, The wall is at L to the floor. 
J_ = perpendicular, The pole stands J_ to the base. 
I I = parallel, The shafts of two spur gears are | | . 
= degree (circular arc or thermometer). The 
temperature was 8o° to-day, or It is a 30 angle. 
' = minutes or feet, It moved 6' in 30' time. 
" = seconds or inches, It moved 3.125" in 5" time, 
h.p. = horse power, The engine develops 25 h.p. 
fo.h.p. = brake horse power, That gas engine develops 

12 b.h.p. 
i.h.p. = indicated horse power, 15 is the actual h.p. 

developed in the cylinder. 
B.t.u. = British thermal unit, That lb. of coal contains 
14000 B.t.u. 
kw. = kilowatt or 1000 watts, The electric generator 
develops 90 kw. 
m.e.p. = mean effective pressure, The m.e.p. is about 

80 lbs. 
r.p.m. = revolutions per minute, The pulley runs 300 

r.p.m. 
f.p.m. = feet per minute, The piston travels 1500 f.p.m. 

Notation and Numeration 

Arithmetic is the science and application of numbers. 

A Number is a unit or collection of units. 

An Integer or whole number is composed of whole units 
only. 

A Concrete Number is one applied to any particular thing. 
Example: 5 boys, 6 men, 25 bolts. 



NOTATION AND NUMERATION 3 

An Abstract Number is one not applied to any particular 
thing. Example: 5, 6, 25. 

Notation is the art of expressing number by figures or 
letters. Example: XII = 12, 5 = five. 

Numeration is the art of reading numbers which have 
been expressed. Example: 545 is read — five hundred forty 
five. 

The Arabic Notation is the method of expressing numbers 
by figures. Example: o = naught or cypher, 1 = one, 
2 = two. 

The location of a figure in a number denotes its value. 
Example: 2 alone = two, but in 21 the value of the 2 is 
increased by its location. 

In reading figures, the first one on the right is units and 
then in order come, tens of units, hundreds of units, thous- 
ands, tens of thousands, hundreds of thousands, millions, 
tens of millions, hundreds of millions, billions, tens of billions, 
hundreds of billions, trillions, tens of trillions, hundreds of 
trillions, etc. Example: 845624891623125. 

It is customary to separate numbers into three figure 
groups by commas. Example: 845,624,891,623,125. 

Rule for Numeration or Reading. — Begin at the left and 
read each period containing one or more figures as if it stood 
alone, adding its name: thus, the above number is read 
845 trillion, 624 billion, 891 million, 623 thousand, 125. 

EXERCISES 

1. Write a seven figure number. 

2. Write an integer. 

3. Give an example of a concrete number. 

4. Give an example of an abstract number. 

5. Express by Arabic notation: eighty-five. 

6. Write out in words: 846,762,845,967,843. 

7. Why does the 9 in 95 represent a greater value than the 9 in 59? 

8. Separate into periods: 1648432. 



INDUSTRIAL MATHEMATICS 



Addition 



Addition, subtraction, multiplication and division are the 
most important processes necessary in mathematical calcu- 
lation. 

Addition is the process of finding the sum of two or more 
numbers. The sign + indicates that addition is to be per- 
formed and is read plus. Example: 2 + 5 = 7. 

The sign of equality is = and is used thus: 2 + 5 = 7. 

In adding concrete numbers, they must be of the same 
denomination. 

Rule. — In adding, place the numbers in a vertical column 
with units under units, tens under tens, etc. Place the right 
hand figure of the sum of the right hand column under units, 
and if there be two figures in this sum, the left hand one is 
carried and added to tens column. Example: 

843 
246 
172 



1261 = (Ans.) 

Subtraction 

Subtraction is the process of taking one number from 
another. Example: 12 — 5 = 7. 

The Minuend is the number from which the other is to be 
taken. Example: 12 — 5 = 7; 12 is the minuend. 

The Subtrahend is the number which is to be taken from 
the minuend. Example: 12 — 5 = 7; 5 is the subtrahend. 

The Remainder is what is left after the subtrahend has 
been taken from the minuend. Example: 12 — 5 = 7:7 is 
the remainder. 

The (— ) sign (called minus) indicates that subtraction is 
to be performed and that the number following it is to be 
taken from the number before it. 



ADDITION, SUBTRACTION, MULTIPLICATION, DIVISION 5 

Rule. — In subtraction, the units must fall under units, 
the tens under tens, etc. Subtract the right hand figure of 
the subtrahend from the right hand figure of the minuend 
and place the remainder directly below. If necessary, 
borrow of the next left hand figure of the minuend to make 
this possible. Then subtract from the remaining minuend 
next. The sum of the subtrahend and remainder should 
always be equal to the minuend. Example: 

4625 minuend 
— 3287 subtrahend 

1338 remainder 
Proof: + 3287 subtrahend 
= 4625 minuend 

Multiplication 

Multiplication is the process of taking or increasing one 
number a certain numbers of times. 

The Multiplier is the number which shows how many 
times the other number is to be taken. Example: 5X8 
= 40; 5 is the multiplier. 

The Multiplicand is the number which is to be taken a 
certain number of times. Example: 5 X 8 = 40; 8 is the 
multiplicand. 

The Product is the result obtained by taking one number 
a certain number of times. Example: 5 X 8 = 40; 40 is 
the product. 

The X sign (called times) indicates that multiplication is 
to be performed and that the number following is to be taken 
as many times as there are units in the number before. 
Example: 5 X 8 = 40. 

Rule. — In multiplication, the units in the multiplier are 
placed under the units in the multiplicand, and the right 
hand figure of the product placed directly below the other 
2 



6 INDUSTRIAL MATHEMATICS 

right hand figures. Each figure of the multiplicand, be- 
ginning at the right, is multiplied by each figure of the 
multiplier and the right hand figure of each partial product 
is placed in turn directly under the figure used as multiplier. 
Partial products are placed on different lines. The sum of 
the partial products will equal the required product. 

Example: 842 multiplicand 
X 245 multiplier 

4210 
3368 
1684 
206290 product 

Division 

Division is the process of determining how many times 
one number is contained in another. Example: 12 -s- 4 = 3, 

12/4 = 3- 

The Dividend is the number to be divided. Example: 
12-7-4 = 3; 12 is the dividend. 

The Divisor is the number by which we divide. Example: 
12-7-4 = 3; 4 is the divisor. 

The Quotient shows how many times the divisor is con- 
tained in the dividend. Example: 12 4- 4 = 3; 3 is the 
quotient. 

The -7- sign indicates that division is to be performed and 
that the number before the sign is to be divided by the 
number following the sign. 

Short Division is used where the divisor contains but one 
figure. Example: 144 -4-8 = 18, 144/8 = 18. 

Long Division is used where the divisor contains more 
than one figure. Example: 378 -s- 14 = 27. 

Rule: Short Division. — Place the divisor at the left of 
the dividend separated by a line and draw a line under the 



ADDITION, SUBTRACTION, MULTIPLICATION, DIVISION 7 

dividend. Try the divisor into the first or first two figures 
of the dividend, as is necessary, and place the quotient under 
the line. If the divisor does not go an even number of times, 
the remainder is prefixed to the next figure in the dividend 
and the process repeated. If there is a final remainder, the 
answer becomes a mixed number. 

Example: 29757 "*" 7 == 4 2 5i- Solution: [ 729757 

4251 

Rule : Long Division. — Place the divisor at the left of the 
dividend, separated by a line, and place the quotient either 
above or to the right of the dividend. Try the divisor into 
the first group of figures which gives a number larger than 
the divisor, place the first figure of the quotient above the 
dividend and (after multiplying it by the divisor) place this 
product below the figures divided into and subtract. The 
remainder prefixed to the next figure brought down, forms 
the new trial dividend. Repeat until all figures of the 
dividend are brought down. 

Example 1, 1 041 741 -v- 243 = 4287. 

Solution: 4287 = (Ans.) 

243 1 1041741 
972 
697 
486 
2114 
1944 
1701 
1701 



8 



INDUSTRIAL MATHEMATICS 



Example 2, 978804 -f- 243 = 4028. 

Solution: 



4028 







243 


I 978804 
972 

68 

00 

680 

486 

1944 

1944 






EXERCISES 




I. 


(a) Add 


842658, 36541. 274, 896 






(b) " 


4657, 125, 84 




2. 


(a) Add 


32689, 54, 97, 125, 18 






(&) " 


63, 762, 81, 91, 5000 




3- 


(a) Add 


666, 83, 12, 6, 7 






(&) " 


5589, 664, 98, 54376 




4- 


(a) Subtract 


864 from 86945 






(&) 


123 " 465 




5- 


(a) Subtract 


64 from 99 






(b) 


126 " 465 




6. 


(a) Subtract 


10 from 1 0000 






(b) 


986 " 2245 




7- 


(a) Multiply 


84 by 84 






(&) 


123 " 17 




8. 


(a) Multiply 


164 by 140 






(&) 


14000 " 7 


■s 


9- 


(a) Multiply 


16845 by 93 






(6) 


2345 " 3456 




0. 


(a) Divide 


20232 by 24 






(b) " 


8262 " 51 




I. 


(a) Divide 


20539 by 893 






(&) " < 


5880635 M 107 




.2. 


(a) Divide 


140010 by 13 






(&) " J 


5001613 " 4957 





CANCELLATION, LEAST COMMON MULTIPLES 9 

Cancellation 

Cancellation is a short method for performing multipli- 
cation and division, applicable in some cases. 

Rule. — In cancellation, all the dividends are placed above 
the line, all the divisors below, and common factors from 
each taken out before multiplying. Example: Divide the 
product of 8, 16 and 12 by the product of 8, 4 and 3. 

Solution: 4 4 



£X4 X i 



= 4 X 4 = 16. (Ans.) 



Least Common Multiples 

A Multiple of a number is a quantity which can be divided 
evenly by the number. 

The Least Common Multiple of two or more numbers is 
the least number which can be divided evenly by both of 
them. 

Rule. — The least common multiple of two or more numbers 
is found by dividing each number into its smallest factors 
and then taking each factor the greatest number of times 
it is found in any one number. Example: L.C.M. of 12 
and 15. 

Solution: 12 = 2 X 2 X 3 

15=3X5 

2X2X3X5 = 60 L.C.M. (Ans.) 

EXERCISES 

1. Divide the product of 48, 39 and 25 by the product of 5, 6 and 13. 

2. Divide the product of 45, 64 and 49 by the product of 7, 9 and 8. 

3. (40 X 150 X 10) -7- (100 X 25) =? 

4. (a) Find L.C.M. of 18, 14 and 12. 
(b) " " " 12, 16 and 4. 

5. (a) Find L.C.M. of 3, 4 and 5. 
(b) " " " 81 and 9. 



10 INDUSTRIAL MATHEMATICS 

6. (a) Find L.C.M. of 72, 3, 9, 2 and 6. 
(&) " " " 9 and 10. 

7. (a) Find L.C.M. of 1, 2, 3, 4 and 5. 
(b) " " " 2, 6, 9, s, 4 and 3. 

Common Fractions 

A Fraction of any thing is a part of it. Example: J = one 
half or f = two thirds. 

The Denominator of a fraction is the number below the 
line and it indicates into how many equal parts the thing is 
to be divided. Example: f ; 8 is the denominator. 

The Numerator is the number above the line and it indi- 
cates how many of the equal parts are to be taken. Ex- 
ample: J; 1 is the numerator. 

A Common Fraction is one whose denominator is not 
10, 100, 1000, etc. Example: J, f, f. 

A Proper Fraction is one whose numerator is less than the 
denominator. Example: f. 

An Improper Fraction is one whose numerator is greater 
than the denominator. Example: |-. 

A Simple Fraction is one not connected with another. 
Example: f. 

A Compound Fraction is a fraction of a fraction. Ex- 
ample, § of f . 

A Complex Fraction is one having a fraction in both or at 

2 1 
least one of its terms. Example:- or ~. 

3 3 

The line drawn between the numerator and denominator 
of a fraction indicates division, or that the numerator is to 
be divided by the denominator. Example: f = 3 -f- 4. 

A Mixed Number consists of a whole number and a 
fraction written together. Example: 3 J or I2J. 

Rule. — To reduce a mixed number to an improper frac- 
tion, multiply the whole number by the denominator and add 



ADDITION, ETC., OF FRACTIONS II 

the numerator, so as to form a new numerator. Example: 

2 

Rule. — To reduce an improper fraction to a mixed number, 
divide the numerator by the denominator. The quotient 
will be the whole number and the remainder placed over the 

divisor will be the fraction. Example:- = 3J. Solution: 

2 

7^2 = 3 and 1 remainder. 

Rule. — To reduce a whole number to a fraction, simply 

multiply the whole number by the denominator of the 

5X4 20 

fraction desired. Example: 5 to 4ths = = — . 

4 4 

Rule. — The value of a fraction is not changed by multi- 
plying or dividing both terms by the same number. 

2X2 4 2 4 -j- 2 2 4 

Example: = - or — . = — or — . 

3X2 6 3 6-7-2 3 6 

Addition of Fractions 

Rule. — In adding fractions, the least common multiple 
of all of the denominators must first be determined, and 
then all fractions changed in form to this denominator. 

235 28 

Example: Add -,- and - L. CM. of 3, 4 and 6 = 12. - = — 
346 3 12' 

3 9 5 10 8 9 10 27 3 1 
-=— ,- = — . —+— + — = — = 2— or 2- (Ans.). 

4 12 6 12 12 12 12 12 12 4 

Where mixed numbers are to be added, the whole numbers 
are added separate and then added to the result. Example: 

ill 6 34 13 1 

2- + 3- + " = 2- + 3- + - = 5- or 6- (Ans.). 
243 12 12 12 12 12 

Rule. — To reduce a fraction to higher terms, divide the 



12 INDUSTRIAL MATHEMATICS 

required denominator by the denominator of the fraction, 
and multiply this quotient by the numerator. Example: 

Reduce | to 8ths; 84-2=4. 4X1=4, or - = - (Ans.). 

2 8 

Subtraction of Fractions 

In subtracting fractions, the denominators must be re- 
duced to a common denominator or L.C.M. Example: 

£_I = i_3 =i(AnsX 

3 2 6 6 6 

In subtracting fractions where the minuend is a mixed 
number with a fraction smaller than the subtrahend, the 
minuend can then be reduced to an improper fraction before 
subtracting. Example: 

1252 5 4 11 5 

2 = ■ -= 1- — - = — or 1- (Ans.). 

2323 666 6 

Multiplication of Fractions 

In multiplication of fractions, reduction to common de- 
nominators is not necessary. Example: J X f = f (Ans.). 

Rule. — To multiply a fraction by a whole number, multiply 
the numerator or divide the denominator by the whole 

4 20 

number. Example: - X 5 = — = 4 (Ans.). 

5 5 

Rule. — To mult'p y two or more fract'ons together, 
multiply all the numerators for a new numerator and all the 
denominators for a new denominator and then reduce to ' 
lowest terms. Example: 

1 X - X i = - (Ans.); i X -X 1 = -- = - (Ans.). 

2 4 5 40 3 8 9 216 27 

Rule. — To multiply two or more fractions and mixed 



ADDITION, ETC., OF FRACTIONS 1 3 

numbers together, the mixed number can first be reduced 
to improper fractions and then the preceding rule be used. 

i i 3 9 27 3 

Example: i-X2- = -X~ = — or 3 - (Ans.). 
2 4248 8 

Rule. — To multiply fractions by cancellation, first reduce 
any mixed numbers to improper fractions and then cancel 
any factors which may be found in both the numerator and 

1 i i 1 
denominator. Example: - X ~ X - = ~ (Ans.). 

Division of Fractions 

In the division of fractions, reduction to a common de- 
nominator is not necessary. 

Rule. — To divide one fraction by another, first reduce 
any mixed numbers to improper fract ons, th n nv.rt the 
divisor and proceed as in multiplication. Example: 

3 . I 3 v 4 I2 1 3 x/ * 3 r^ \ 

- .*. - = _ x - = — = 3, also - X - = -= 3 (Ans.). 

44414 ^ 1 .1 

EXERCISES 

1. Reduce to mixed numbers: \ -, 2 7 2 , ¥> ¥. H~> W. ^i* Iff' 

12 3 13 

T2T» T2- 

2. Reduce to improper fractions: s|, 8^, 6§, I2f, 14^, 6|, n\\, 
49|f, 888|ff, i 25t 9 t . 

3. Reduce to simple fractions: \ of £, |- of T V re of f, f of T f , 

3 _f 5 2 r _9_ lr 5 8 _.f 5 121 „f 389 862 „f I 

4 Ot 9 , 3 Ot 16 , 2 Ot ?. T7 0t T9' T?6 0t ¥6 3' T?3 0t 2- 

4. Reduce to simple fractions: 

62 125 18 14 1 1 1 13 8 

fijf T3 2" H H I J II J5 ¥ 

1 ■ 8 ' «. ' II* 8' 5' 1 ' 13' 2_' 

2 ¥ 9 1 6 ¥ 4 ll 3 

5. Reduce to fractions: 7 to 4ths, 18 to halves, 12 to 3rds, 8 to I2ths, 

5 to ioths, 6 to 3rds, 4 to 8ths, 6 to 7ths, 4 to Sths, 17 to I3ths. 

6. Reduce to lowest terms: T |, T \, T |f, - 6 <f, 14^-, i8 T \, i24 T |f, 

7. Reduce to higher terms: f to 64ths, § to 48ths, -3% to i28ths, 
r| to 64ths, M to i6oths, i| to 8ths, 8| to 24ths, V to 6ths. 



14 INDUSTRIAL MATHEMATICS 

8. Reduce to least common denominator: |, f, ■§: also f, f, -f: 
also 8|, 3-|, |: also £■, 5%: also |, \, a|: also 5$, 3 J, ^ :also f, 
S^J, 1^: also 8 J, a|, ^: also 8|, &. 

9. (a) Add I, f, 2J, sf. 

(6) " 2f, 2 S |, i8f, if. 

10. (a) Add 865, i i 
(6) " 4 iT 5 3. 4ItV 

11. (a) Add 22|, 22j, 22^, 22^. 
(&) " 2l|, 489T6, 2f. 

12. (a) Subtract 32 1 from 40. 
(&) " 18* " 19J. 

13. (a) Subtract 22 f from 29^. 
W " 2^ " 3-3^. 

14. (a) Subtract 465^3 from 491^. 
(b) " ioof " 9862!. 

15. (a) Multiply | by f . 

(6) " i X 2\ X si X A. 

16. (a) Multiply | by f. 

(6) " 2^X4^X5|X4i 

17. (a) Multiply 4§ X iM X | X 2j. 
(&) " Si X 19I X Sf X 8f, 

18. (a) Divide f by f . 
(6) " 2\ by i|. 

19. (a) Divide ij by 2^. 
(6) " 8 4 | by i 7 f. 

20. (a) Divide 265! by 4§. 
(6) " 489I by 93!. 

21. If a machinist finds five different sizes on a shaft, each one being 
respectively 4!", 3I", 1", f" and -fa" long, how long is the entire shaft 
in inches? 

22. What is the combined height of six blocks of steel placed upon 
each other, the sizes of each being \", f", &", \", &" and if"? 

23. How many hours will it take to do all the operations on 5 shafts » 
if the centering on each requires \ hour, the rough turning \ hour, the 
finish turning \ hour and the threading \ hour? 

24. How long will the front bearing on a lathe spindle be, if the 
entire length is 2 9/24 ft., the nose is \ ft., the space for the cone is 
1 5/24 ft., the rear bearing is \ ft., and the remaining end is 7/24 ft.? 

25. What will it cost to produce 12 gears at 30 cents per hour for 
^abor if the first operation requires i| hours, the second if hours, and 
the third 2? hours? 



DECIMAL FRACTIONS 1 5 

26. If S hangers are used to support a line shaft 42^ ft. long, what is 
the distance between the hangers? 

27. A man purchased a 2/5 interest in a factory and after giving away 
^ of this, divided the remainder among his three sons. What part of 
the original factory does each own? 

28. How many posts 8| ft. long can be cut from a tree 32 f ft. high? 

29. How many ties, if they are placed 2\ ft. apart, will be required 
for 3265 ft. of track? 

30. If a pulley is 3I ft. in circumference, how many times will it 
revolve in causing a belt 28§ ft. long, to make one complete travel of 
its length? 

Decimal Fractions 

A decimal number is one containing figures to the right 
of the units place, separated from the unit figure by a decimal 
point or period. Example: 10.2 is a decimal number. 

The figures to the right of the decimal point are read, in 
order, as tenths, hundredths, thousandths, ten thousandths, 
etc., each figure having one-tenth the value of the left hand 
figure preceding. Example: 10.201 = ten and two tenths 
and no hundredths and one thousandth, or ten and two 

201 

hundred one thousandths, or 10 ■ . 

1000 

In some cases, the decimal fraction method is not exactly 
accurate, but for most calculations, it is sufficiently accurate, 
and much more convenient than the use of common fractions. 

The denominator of a decimal fraction may be determined 
by placing the figure 1 under the decimal point and a cipher 
under each figure of the decimal. Example: 

0.465I , , 1.2I , 

r = 465/1000. \ = 1 2/10. 

ioooj 10J 

The figures at the left of the decimal point represent the 
whole number and those at the right of the point the fraction. 

Example: 25.45 =25 . 

100 



1 6 INDUSTRIAL MATHEMATICS 

Ciphers added at the right hand side of a decimal fraction 
do not change its value. Example: 3.25 or 3.250 or 3.2500. 

A common fraction can be reduced to a decimal exactly 
or approximately by dividing the numerator by the denom- 
inator. Example: f = 0.625. 

Rule.' — To reduce a common fraction to a decimal, place 
a point at the right of the numerator and divide the denom- 
inator into this number. If it will not go, add a cipher to 
the right of the point and proceed as in division, adding 
ciphers as required. If after several divisions, the remainder 
does not disappear, it is probably a repeating decimal, and 
the remainder should be dropped. The quotient should be 
pointed off from the right as many places as the number of 
places in the dividend exceed the number in the divisor. 

Example: 5/8 = 8|jvOOO 5 /6 = 6 [ 5-000 

0.625 (Ans.). 0.833 (Ans.). 

Moving the decimal point to the right, multiplies the 
number by (10) for each place moved. Moving to the left, 
divides by (10) for each place moved. Example: 

3.45 X 10 = 34.5 (Ans.). 3.45 -;- 10 = 0.345 (Ans.). 

Addition of Decimals 

In addition or subtraction of decimals, the points should 
always fall under each other in the fraction and in the sum. 

Example: 2.12 

+ 14-3 

16.42 (Ans.). 

Addition is performed the same as in the addition of whole 
numbers. 



ADDITION, ETC., OF DECIMALS 17 

Subtraction of Decimals 

Subtraction is performed the same as in the subtraction of 
whole numbers. Example: 34.620 

- 4-315 
30.305 (Ans.). 

Multiplication of Decimals 

In multiplication of decimals, the points are not required 
to fall under each other and the fractions are placed generally 
so that the right hand figures in the multiplier and multi- 
plicand will fall under each other. Example: 

3.45 X 2.5 = 3.45 
X 2.5 

1725 
690 

8.625 (Ans.). 

Rule. — In multiplication of decimal fractions, the number 
of places pointed off from the right in the product should 
equal the sum of the places in the multiplicand and multi- 
plier. Example above. 

Division of Decimals 

Rule. — Division of decimal fractions is exactly the same 
as division of numbers, except by the adding of ciphers to 
the right of the decimal point in the dividend, which is 
sometimes necessary. Point off as many places in the 
quotient as the number of places in the dividend exceeds 
the number of places in the divisor. Example: 

2.5 -r- 1.25 = 1-25 2.50' 2.0 (Ans.). 
2.50 

To reduce a decimal fraction to a common fraction, place 



1 8 INDUSTRIAL MATHEMATICS 

the denominator of the decimal below it and reduce to 
lowest terms. Example: 

625 625 -i- 125 5 

0.625 = " or = - (Ans.). 

1000 1000 -f- 125 8 

EXERCISES 



I. 


(a) Add 


3.25, 4.625, 46.865, 


4.25. 




(b) •' 


86.000, 86000.3. 




2. 


(a) Add 


0.125, 125, 1.25. 






(6) " 


4-65793. 46489-73- 




3- 


(a) Add 


2649.37, 0.0000467, 


100.0009 




(b) " 


1004.373, 58.9876, 


1.26578. 


4- 


(a) Subtract 


4.625 from 84. 






(6) 


.86497 from 1. 





5. (a) Subtract 3.460009 from 4.00. 
(b) " 0.006387 from 0.06487. 

6. (a) Subtract 13.399 from 14.01. 
(&) " 133-99 from 407.387- 

7. (a) Multiply 34-65 by 3.6. 
(6) " 287.03 by 48.0 

8. (a) Multiply 465 by 1.5. 
(&) " 0.83 by 0.83. 

9. (a) Multiply 0.625 by 0.75. 
(6) " 4.87 by 27.4. 

10. (a) Divide 8.463 by 4.65 (to 3 places). 
(&) " 48.43 by 4.5 (to 3 places). 

11. (a) Divide 165 by 0.165. 
(6) " 84003 by 0.0006. 

12. (a) Divide 0.00006 by 3. 
(b) " 9-63 by 6. 

13. (a) Reduce to decimal fractions: 1/64, 1/8. 
(b) " " " " 5/7. 3/8. 

14. (a) Reduce to decimal fractions: 7/64. 8/9. 
(b) " " " " 2/3. 4 5/6. 

15. (a) Reduce to common fractions: .5, .75. 

(&) " " " 2.625, 4.40625. 

16. (a) Reduce to common fractions: 3.546875, 7.1875. 
(b) " " " 25.046875. 1-375- 

17. What will 17.5 yards of canvas cost at $0.43 per yard? 



PERCENTAGE 1 9 

18. What will 25 castings, weighing 0.5 lbs. each, cost at $0.30 per lb.? 

19. What rate must be charged for butter so that 12.5 lbs. will net 
$4.0625? 

20. What will be the weight of 33.25 cu. in. of iron if it weighs 0.25 
lbs. per cu. in.? 

Percentage 

1 

Percentage means per hundred = 1%. 

100 

Percent means a certain number of hundredths. 

The percent sign is (%) and represents hundredths. 

10 

Example: 10% = or 0.10 or 0.1. 

100 

The Base is the number on which the percent is calculated. 

Example: 5% of 12 = 0.6; 12 is the base. 

The Rate percent is the number of hundredths of the base. 

Example: 5% of 12 = 0.6; 5% is the rate. 

The Percentage is the result obtained by taking a certain 

number of hundredths of the base. Example: 5% of 12 

= 0.6; 0.6 is the percentage. 

The Amount is the sum of the percentage and base. 

Example: 5% of 12 = 0.6; 0.6 + 12 = 12.6, amount. 

The Difference is the difference between the percentage 

and the base. Example: 5% of 12 = 0.6; 12 — 0.6 = 11.4, 

difference. 

Rules: Base X rate = percentage. 

Percentage 4- rate = base. 

Percentage -f- base = rate. 

Base X (1 -f rate) = amount. 

Amount -*- (1 -f- rate) = base. 

Base X (1 — rate) = difference. 

EXERCISES 

1 . Find 3% of $800.00. 4. Find what percent 26 is of 326. 

2. Find 45% of 200. 5. Find what percent 3 is of 2. 

3. Find 7^% of 20 gallons. 6. Find what percent 45 is of 45. 



20 INDUSTRIAL MATHEMATICS 

7. What number of gears must be ordered from foundry if 5% are 
poor castings, to have 250 good ones? 

8. What number if increased by 8% of itself will give 400? 

9. What amount of money must be invested at 5% to amount to 
$300 in one year's time? 

10. If 30% is lost through friction, what power will be transmitted 
from a mechanical device if 24h.p. is supplied? 

WEIGHTS AND MEASURES 
Table of Linear Measure 

12 inches = 1 foot 

3 feet or 36 inches = 1 yard 

5 \ yards or 16^ feet = 1 rod 

40 rods = 1 furlong 
•8 furlongs or 320 rods, 

or 1760 yards or 5280 feet = 1 mile 

Table of Square Measure 

144 square inches = 1 square foot 

9 square feet = 1 square yard 
30! square yards 

or 272! square feet = 1 square rod 
160 square rods or 4840 square 

yards or 43560 square feet = 1 acre 

640 acres = 1 square mile 

Table of Cubic Measure 

1728 cubic inches = 1 cubic foot 

27 cubic feet = 1 cubic yard 

128 cubic feet = 1 cord 

Table of Avoirdupois 
435-5 grains = 1 ounce 

16 ounces or 7000 grains = 1 pound 

100 pounds = 1 hundredweight 

20 cwt. or 2000 pounds = 1 ton 

2240 pounds = 1 long ton 

Table of Troy Weight 

24 grains = 1 pennyweight 

20 pennyweights or 480 grains = 1 ounce 
12 ounces or 5760 grains = 1 pound 



WEIGHTS AND MEASURES 21 

Table of Apothecaries Weight 
20 grains = i scruple 

3 scruples or 6o grains = I dram 
8 drams or 480 grains = 1 ounce 
12 ounces or 5760 grains = 1 pound 

The grain is the same in Avoirdupois, Troy and Apothecaries weights. 

Table of Liquid Measure 

4 gills = 1 pint 
2 pints or 8 gills = 1 quart 

4 quarts or 231 cubic inches = 1 U. S. gallon 
31^ gallon = 1 barrel 

2 barrels = 1 hogshead 

Table of Dry Measure 

2 pints = 1 quart 
8 quarts or 16 pints = 1 peck 
4 pecks 

or 2150.42 cubic inches = 1 bushel 

Table of Angular Measure 
60 seconds = 1 minute 

60 minutes = 1 degree 

90 degrees = 1 quadrant 

360 degrees = 1 circle 

Table of Time Measure 

60 seconds = 1 minute 

60 minutes or 3600 seconds = 1 hour 

24 hours or 1440 minutes = 1 day 

7 days =1 week 

52 weeks or 365^ days = 1 year 

100 years = 1 century 

Miscellaneous Table 
1 cubic foot =7.48 gallons 

1 inch = 25.4 millimeters 

231 cubic inches = 1 U. S. gallon. 

660 feet = 1 furlong 

3 miles = 1 league 

23 feet = 1 military pace 



22 



INDUSTRIAL MATHEMATICS 



2 yards 
24! cubic feet 
2150.42 cu. in. 
5760 grains 

7000 grains 
144 cu. in. 
39.37 inches 
62.5 pounds 
264.2 gallons 
1550 sq. in. 
3.168 grains 
15-432 grains 
746 watts 
1000 watts 
0.26 lbs. cast iron 
0.283 lbs. steel 
0.092 lbs. aluminum 
0.300 lbs. brass 
2545 B.t.u. 



= 1 fathom 
= 1 perch 
= 1 U. S. bushel 
= 1 pound Troy or 
Apothecaries' 
= 1 pound Avoirdupois 
= 1 board foot 
= 1 meter 

= 1 cubic foot of water at 62 ° F. 
= 1 cubic meter 
— - 1 sq. meter 
= 1 carat 
= 1 gram 
= 1 h.p. 
= 1 k.w. 
= 1 cubic inch 
= 1 cubic inch 
= 1 cubic inch 
= 1 cubic inch 
= 1 h.p. hour 



Rule. — To change a quantity to a higher denomination, 
divide the quantity by the number of parts of the quantity 
required to make one of the higher denomination. Example: 
Change 15 inches to feet. 15 -f- 12 = ij ft. (Ans.). 

Rule. — To change a quantity to a lower denomination, 
multiply the quantity by the number of parts of the lower 
denomination required to make one of the quantity. Ex- 
amp e: Change 4 feet to inches. 4 X 12 = 48 inches 

(Ans.). 

EXERCISES 

1. Reduce 12 rods, 4 yards to inches. 

2. Reduce 200 inches to higher denomination. 

3. Reduce ij.ooo feet to miles and feet. 

4. Reduce 95 furlongs to miles and feet. 

5. Reduce 17I yards to feet and inches. 

6. Find area of field in acres which is 3 miles by 2 miles. 

7. Find number of square inches of surface on a box 3" X 4" X 5"- 

8. Find number of sq. yds. and sq. ft. in tract of land 480 X 500 ft. 



WEIGHTS AND MEASURES 23 

9. Reduce 1225 sq. yds. to sq. ft. 

10. Reduce 127,050 sq. yds. to sq. rds. 

11. What will be the cubic contents of a box 14" X 92" X 71" in 
cu. ft.? 

12. Reduce a cord of wood to cubic inches. 

13. Reduce 88 cu. yds. to cu. ft. 

14. Reduce 145 cu. ft. to cu. in. 

15. How many grains in 14 lbs. 12 oz. (Avoirdupois)? 

16. How many ounces in a ton? 

17. How many tons and ounces in 3,000,000 ounces? 

18. How many hundred weight in 5 tons? 
19.- Reduce to ounces, 3 tons, 4 cwt., 23 lbs. 

20. Reduce to pwt., 400 grains. 

21. Reduce to Troy ounces and grains, 50,000 grains. 

22. Reduce to Troy pounds and ounces, 75 ounces. 

23. Reduce to Troy ounces, 5 Troy pounds and 200 grains. 

24. Reduce to grains, 200 Troy lbs. 

25. Reduce to scruples, 200 grains. 

26. Reduce to drams, 1500 grains. 

27. Reduce to Apothecary lbs. and oz., 43,000 grains. 

28. Reduce to Apothecary ounces, 7 lbs. 

29. Reduce to Apothecary drams, 8 lbs. 3 oz. 

30. Reduce to barrels, 252 gallons. 

31. Reduce to quarts, 42 pints. 

32. Reduce to gills, 2 barrels. 

33. Reduce to bushels, 64 pints. 

34. Reduce to pints, 64 bushels. 

35. Reduce to gallons and gills, 180 gills. 

36. Reduce to quarts, 18 pecks. 

37. Reduce to pecks, 800 quarts. 

38. Reduce to pints, 40 bushels. 

39. Reduce to seconds, 90 degrees. 

40. Reduce to degrees and minutes, 400 minutes. 

41. Reduce to seconds, 80 degrees and 45 minutes. 

42. Reduce to minutes, 3600 seconds. 

43. Reduce to seconds, 360 degrees. 

44. Reduce to days, 14 years. 

45. Reduce to years, 3287I days. 

46. Reduce to minutes, 45 hours. 

47. Reduce to hours and minutes, 27 days. 

48. Reduce to minutes, 2 years. 



24 INDUSTRIAL MATHEMATICS 

49. Reduce 4s " to millimeters. 

50. Reduce 120 millimeters to inches. 

Ratio and Proportion 

Ratio is the comparative size of two numbers. Example: 
12 to 8 = 12 -2- 8 = ij. 

The sign of ratio is a colon (:). Example: 8 : 4. 

The first term of a ratio is the Antecedent. 

The second term of a ratio is the Consequent. 

The two terms of a ratio must be like numbers, as $8 : $4, 
5 men to 3 men. 

A Compound Ratio is the product of two or more simple 

ratios. Example: { \ \ as 10 : (?). 

A compound ratio is reduced to a simple one by multi- 
plying the antecedents together for a new antecedent and 
the consequents for a new consequent. Example: 

4 ;^ : : 10: (?) = 20:6 : : 10: (?). 

A Proportion is an equality of two ratios. Example: 
12 is to 6 as 4 is to 2. 

The sign of proportion is the double colon (: :). Example: 
14 : 7 : :4 : 2. 

The first and fourth terms of a proportionate are called 
the Extremes and the second and third are called the Means. 
Example: 14 : 7 : : 4 : 2. 
E M M E 

In a Direct Proportion both couplets are direct ratios. 
Example: 5 men : 8 men : : $30 : $48. Where the amount 
8 men would earn is required. 

An Inverse Proportion requires reversing one of the 
couplets. Example 8 : 5 : : 30 days : 18 days. Where 30 
days is the time required for 5 men to do a job and the time 
required for 8 men to do it is wanted. 



RATIO AND PROPORTION 25 

When 3 numbers are in a proportion so that the 1st is to 
the 2d as the 2d is to the 3d, the 2d number is a Mean Pro-^ 
portion between the 1st and 3d. Example: 2-4-8. 

Rules. — The product of means = product of extremes; 

The product of mean ■*■ extreme = other extreme; 
The product of extremes -*- mean = other mean. 

The cause and effect method can be used to advantage in 
compound proportion by placing all the causes on one side 
of the : : and all the effects on the opposite side. Example: 
If 1 1 men can assemble 45 motors in 6 days of 10 hours each, 
how many men will it take to assemble 81 motors in 12 days 
of 11 hours each? 

1st cause 2d cause 1st effect 2d effect 



11 men ? men 

■ 6 days r : -j 12 days 
10 hours 11 hours 



H : : 45 motors : 81 motors, 
extremes 11 X 6 X 10 X 81 



means ? X 12 X 11 X 45 



= 9 men = (Ans.), 



EXERCISES 

1. If 3 men can drill 200 pieces in one hour, how many can 8 men 
drill? 

2. If one gallon of oil will last a department 14 days of 10 hours 
each, how many gallons of oil will be used in 6 weeks of 55 hours each? 

3. If the payroll each week is $500 for 15 men, at this ratio what 
would it be for 40 men? 

4. If a Frankfort gas furnace consumes 5 ft. of gas per minute, 
when heating Novo Superior steel, what will it cost to operate the 
furnace z\ hours at 60 cents per 1000 ft.? 

5. If we have a six pitch lead screw and a job of 10 pitch, what gear 
would be used with a 40 to cut this thread? 

6. If 12 qts. of oil will last an auto owner 40 days, how long would 
it last four men? 

7. If 48 bars of stock will last 8 screw machines 12 days, how long 
will they last 6 machines? 



26 INDUSTRIAL MATHEMATICS 

8. If it takes 20 days for 30 men to assemble a lot of automobiles, 
how long will it take 45 men to do it? 

9. If 287,000 pieces a day can be produced by operating 7 dies at 
one time, how many days would be required to finish this many if 3 dies 
only were used? 

10. If 5280 men can do a job in 73 days, how long will it take 427 
men to do it? 

11. If 40 men in 15 days, working 10 hours a day can make parts for 
5 gas engines, how many men in 20 days working 8 hours a day would 
be required to make parts for 12 gas engines? 

12. With tool steel at 25 cents a lb., it costs $180 a month to supply 
lathe tools in 3 departments and reamers in 20 departments. At this 
rate what will be the monthly cost if tool steel advances to 30 cents a lb. 
and three lathe and 8 drilling departments are added? 

13. If it costs 30 cents to plate 240 pieces of sheet metal 4" long and 
2" wide, what will it cost to plate 180 pieces 14" long and \" wide? 

14. If 3 men working 6 hours a day can paint the side of a building 
200 ft. long and 80 ft. high in 10 days, how long would it take 14 men 
working 8 hours a day to paint the side of a building 300 ft. long and 
100 ft. high? 

15. If a \" piece of metal cut from the end of a 3" diameter bar 
weighs 14 oz., what would be the weight of a \" piece cut from a round 
bar of the same material but 8" in diameter? 

Taper Calculations 

A piece is said to taper when there is a gradual and uniform 
increase or decrease in its diameter or thickness. Examples: 

A lathe center [ r > or a wedge 

The Amount of Taper is expressed as a certain number of 
inches or parts of an inch per foot and indicates a variation 
in diameter or thickness of that amount in twelve inches of 
length. 

The Standard Tapers used in shop work and their approx- 
imate tapers per foot are the Morse §" per ft.; Brown & 
Sharpe \" per ft.; Jarno 0.6" per ft.; Sellers and Pipe taper 
f " per ft. ; Pratt & Whitney pins \" per ft. 



TAPER CALCULATION 



27 



The taper on the B. & S. No. 10 is 0.516" per ft., and on 
the Morse No. 1 to No. 6 it varies as much as 0.025 t™ m 
§" per ft. 

Taper Turning on the lathe is sometimes accomplished by 
setting over the tail-stock and when this is done the taper 
per unit length as well as the length of the piece is required. 

Rule. — Set the tail-stock over half the amount of the taper 
per ft. times the length of the work in feet. Example: To 
cut a Jarno taper on a piece 9" long; Solution No. 1 : \ of 
0.6 = 3/10"; 3/10 X 3/4 = 9/4o" or 0.225" set over = 
(Ans.); Solution No. 2: 0.6/2" : 12" ::(?): 9" = 0.3" 




EXERCISES 
1. How far must the tail-stock of a lathe be set over to cut a B. 
& S. taper on piece 2\" long? 



2. 


Jarno taper 


3n 


piece 6" long. 


3. 


Morse No. 2 


of 


0.602" 


on piece 2" Ion 


4- 


Morse No. 4 


of 


0.623" 


" " 5" ' 


5. 


Sellers 






.« « S r, . 


6. 


P. & W. pin 






" " 3" ' 


7- 


Pipe tap 






" " 4" ' 


8. 


Jarno 






" " 14" ' 


9- 


B. & S. No. 


10 




•« ■■ 6|" " 



10. If a taper piece measures 0.3875 at one point and 0.4365" at a 
point 2.\" away, how far should the tail stock be set over to cut this 
same taper on a piece 8" long? 



28 • INDUSTRIAL MATHEMATICS 

Interest 

Interest is money paid for the use of money. 

Percent means hundredths. Example: 6% = .06 = 6/100. 

The Rate of Interest is the rate percent per annum of the 
principal paid for the use of money. Example: $100 at 6% 
for 3 years. — Rate of interest = 6%. 

The sum loaned is called the Principal. Example: $100 
at 6% for 3 years. $100 is the principal. 

The principal plus the interest is called the Amount. 
Example: $100 at 6% simple interest for 3 years = $100 
+ $18 = $118 amount. 

Simple Interest is interest on the principal only. 

Annual Interest is simple interest upon the principal and 
upon any interest overdue. 

Compound Interest is interest upon the principal and its 
unpaid interest combined at regular stated intervals. These 
intervals may be annual, semi-annual or quarterly. 

Rule. — The Simple Interest on a sum of money is found by 
multiplying together the principal, the rate and the time in 
years. Example: $100 at 5% for 3J years. 

$100 = principal 

.05 = rate 
$5.00 = interest for 1 year 



$17.50 = interest for 3^ years = (Ans.) 

Rule. — The Annual Interest on a sum of money is found 
by adding together the interest on the principal for the entire 
length of time, and the interest on each year's interest for 
the time it is unpaid. 
Example: $100 at 5% for 3! years. 
1st year's interest = $5. 

The interest for the first year remains unpaid for 
2§ years, the interest for the second year ij 



INTEREST 



2 9 



years, and the interest for the third year \ years. 
Therefore, the unpaid interest draws interest 
for 2\ years, \\ years and \ year, or 4J years, 
and the interest upon $5.00 for that time is 
($5.00 at 5% for 4! years) = $1,125. 
The entire interest due is $17.50 + $1,125 
= $18,625 (Ans.). 

Rule. — The amount at Compound Interest on a sum of 
money is found by adding the simple interest to the principal 
at regular stated intervals, and using this amount as a new 
principal. Example: $100 at 5% for 3J years. 

$100 = principal. 

5 = int. for 1st yr. at 5%. 
I0 5 = prin. for 2d yr. 

5-25 = int. for 2d yr. at 5%. 
110.25 = prin. for 3d yr. 

5-5 1 = int. for 3d yr. 
115.76 = prin. for 4th yr. 

2.89 = int. for 6 mths. 
1 18.65 ~ amt. for 3! yrs. at 5%. 
100.00 = original principal. 
18.65 = compound interest. 



1. 
2. 
3- 
4- 
5- 
6. 
7- 
8. 
9- 
10. 



EXERCISES 

Find simple interest on $ 425 for 3 yrs. 4 mo. at 5% 

.. Ig43 '.. 2 u 6 „ « 6% 

" 1500 " 5 " 9 " " 7% 
Find amount at annual interest on $ 500 for 4! yrs. 

1800 " 10 



" compound 



1250 
460 
500 

1800 
800 



3f 

3 

4* 

10 

1\ 



at 3%. 
" 6%. 
" 5%. 
" 4%- 
" 3%. 
" 7%. 
" 6%. 



30 INDUSTRIAL MATHEMATICS 

Pulley and Gear Diameters 

Where power is transmitted from one shaft to another by 
means of pulleys, gears, belts, chains, etc., the ratio of the 
speeds is the inverse ratio of the diameter of the pulleys or 
gears: or in other words, the pulley diameters vary inversely 
as the speeds vary. Example: A line shaft (Fig. I) turning 
at 240 r.p.m. carries a pulley 12" in diameter connected by a 
belt to a countershaft pulley 8" in diameter; the proportion 
would read to find speed of 8" pulley. 8" : 12" : : 240 
r.p.m. : (?). Q /r 



W 



1 1 



3 120 

it x M0 , ,. , II 

= 360 r.p.m. (Ans.). 



LJ \i 



4 Countershaft M '360 R.PM. 

Fig. I 

The above rule applies to simple gearing also providing 
the diameters are such as to give a whole number of teeth 
to the gears. Example: A gas engine crank shaft (Fig. II) 
turning 2000 r.p.m. has a 24 tooth gear keyed on to it which 
meshes with a 48 tooth gear on the cam shaft. Find the 
speed of cam shaft. 48 teeth : 24 teeth : : 2000 r.p.m. : (?). 



48 T. 




1000 



= 1000 r.p.m. (Ans.). 



2000RPM. ^J_^/000RPA? 
Fig. II 



The Resultant Ratio between the first and last shaft con- 
nected by compound gearing can be found by dividing the 
product of the number of teeth on all the driving gears by 
the product of all the driven gears. Example: Fig. III. 



PULLEY AND GEAR DIAMETERS 



31 



4662 
Driving = /00 X jj</> X jffl X X00 _ 288 
Driven = ^X ^X ^X ^~ 1 
.'. ratio is 288 to 1 (Ans.). 



I SOT. 



2 ST. 




1— 



20 T 



Fig. Ill 



2 ST. 




100 T. 



SOT. 

M 

Driven 



The above rule also applies to a train of shafts and pulleys 
connected by belts and in such cases the pulley diameters 
are used in place of the number of teeth as in gearing. Ex- 
ample Fig. IV. 

2 2 

Drivers = M" X #" 4 . '. , ,. * 

= - . . ratio is 4 to 1 (Ans.). 

Driven = /0" X U" 1 




10QOR.P.M' 



Fig. IV 

EXERCISES 

1. If the front sprocket on a bicycle has 80 teeth and the back one 25 
and the crank shaft is turned 80 r.p.m., how fast will the rear wheel 
go in r.p.m.? 

2. If the line shaft turns 280 r.p.m., and carries a 22" pulley belted 
to a 10" pulley, what will this run in r.p.m.? 

3. An electric motor running at 1500 r.p.m. has a 20 tooth gear on 
the armature shaft, running in mesh with a 180 tooth gear on the driven 
shaft. What is the speed in r.p.m. of the driven shaft. 



32 



INDUSTRIAL MATHEMATICS 



4. Solve for ratio: 

80 T. 30 J. 



nil 



m 



100 T. 



1 



11 iniljfiii % 



T. 2ST. 

MHIIMBMWIlillll 



IlMi ll'llTlll 

01 20T. 



5. Solve for ratio: 

20J. 



18 J. 



Ill 111 1 111)11 111 III 




40 T 20T. 

6. What diameter pulley will be re- 
quired on the line shaft to give 3000 
r. p.m. to a 5" emery wheel if the line 
shaft runs 200 r.p.m. the pulleys from 
countershaft to emery wheel spindle give 
a ratio of 1 to 10 and the pulley belted 
to line shaft is 4" diameter? 

7. Calculate the eight different 
spindle speeds possible on the follow- 
ing 18" back geared lathe, if the coun- 
tershaft runs 300 r.p.m. 



ZOOOR.RM. 


-/O'H 


-8"- 


-■$»- 


-4"- 


300 R.P.M. 


-fc 


-h 














PULLEY AND GEAR DIAMETERS 



33 



8. What will be 
the gear ratio in a 
train of gears if 6 
driver gears of 21 
teeth each 'drive 
6 driven gears of 
63 teeth each? 

9. Calculate the 
eight different 
spindle speeds 
that will be ob- 
tained by the fol- 
lowing sketch of a 
16" back gear 
lathe head. 

10. What is the 
travel forward 
and reverse in 
feet per minute of 
the ram of a 
shaper, with pul- 
leys and gears of 
the size given in 
the sketch? 



e 



220R.RM. 



Uc J 




-2%'Diam.?SJ. 



34 INDUSTRIAL MATHEMATICS 

Square Root, Involution and Evolution 

The Square Root of a quantity is the number which when 
multiplied by itself will produce the quantity. Example: 
V16 = 4, as 4 X 4 = 16. 

When a number is multiplied by itself one or more times, 
the process is called Involution. The number thus multiplied 
is called the Root and the products are called the Powers of 
a number. Example: 4 2 = 16 or 5 2 = 25, This process is 
called involution. 

Any number multiplied by itself once is said to be squared, 
and the product so obtained is called the Square or Second 
Power of the original number. Example: 3* = 9. Thus 9 
is the square or the second power of 3. 

The small figure written near the top of a number, such as 
3 2 , 6 3 is called the Index or Exponent of the number. 

Evolution is the reverse of involution and it means to 
extract or find the root of any given power. Example: 
V16 = 4 or V25 = 5. This process is called evolution. 

The radical sign ( V ) indicates that the square root is to 
be extracted. Example: V16 = 4. 

If a radical be raised to the same power as the index of the 
radical, the radical sign is removed, thus ^2X2X2 = 2 
(Ans.). 

In solving square root problems, the number must be first 
separated into two figure periods by starting from the decimal 
point and working both ways. As 4'83'92.84'6. 



The index 2 need not be 


used, 


as 


■^843.25 = V843.25. 


Example: Extract the .« 


square 


root of 625. = 


2 






2 5 (Ans.). 
V6'2 5 . 


2 






4 


(T.D.) ^ 






225 


_5 






225 


(R.D.) 45 









SQUARE ROOT, INVOLUTION AND EVOLUTION 35 

Example: Extract the square root of 1 19025. = 

345 (Ans.). 



3 


Vn'90'25 


3_ 


9 


60 


2 90 


_4 


2 56 


64 


34 25 


J_ 


34 25 


680 




__5 




685 





Rule. — First point off in two figure periods from decimal 
point, then place square root of first period in answer and 
also at left of problem, next square first figure and subtract 
from period, also bring down next period. Next add first 
division to itself, annex cipher and label it trial divisor (T.D.). 
Next try trial divisor into dividend and place answer in 
answer above radical sign. Next add this number to trial 
divisor for real divisor (R.D.) and after multiplying, sub- 
tract from partial dividend, and bring down next period. 
Next add last number in answer to real divisor, annex one 
cipher and label it trial divisor and continue as before. 

Exceptions to Rule. — (i) If trial divisor is found larger 
than dividend, annex one cipher to it and bring down next 
period in dividend. (2) If trial divisor will go a certain 
number of times, but real divisor will not, it then becomes 
necessary to use a smaller divisor. 

If the quotient is placed above the periods, the decimal 
points will fall under each other. 

EXERCISES 

1. Extract the square root of 246016. 

2. Extract the square root of 132496. 



36 



INDUSTRIAL MATHEMATICS 



3. Extract the square root of 49284. 

4. Extract the square root of 24649. 

5. Extract the square root of 462.846743. 

6. Extract the square root of 1234321. 

7. Find the 9th ppwer of 9. 



8. Find the V 6 X 6 X6. 

9. Find the 7th power of 7. 
10. Find the 2d power of 125. 

EXERCISES 

A. Where B equals the base of a right angle triangle, A equals the 
altitude and H the hypotenuse, the rules A 2 + B 2 = H 2 , H 2 - B 2 = A 2 
and H 2 —A 2 = B 2 can be used for solving for one unknown side. 
(Fig. I.) 




6 *> 
4 



Fig. II 

B. Fig. II shows diagramatically that a right angle triangle B of 4". 
A of 3" will give 5" H. 

1. How long a ladder will reach to the top of a fence 15 ft. high, if 
the ladder is placed 6 ft. away at the bottom? 

2. How long must guy lines be to support a steel stack 125 ft. high 
if the lines are fastened 30 ft. from the top of the stack and the other 
ends are fastened to anchors in the ground, which are located 75 ft. from 
the base of the stack? 

3. How much do I save on a corner by cutting across lots if the 
point where I leave the walk is 350 ft. from the corner 
and when I reach the walk again, it is 425 ft. from the 
corner? 

4. What is the largest size square steel stock that 
will fit into a \" round hole? 

5. What length guy will I have to purchase to 
fasten to the top of a mast 100 ft. high, and to the 
ground 150 ft. away? 




CUBE ROOT 



37 



6. If in boring a jig the diagonal measures 5.124" and the distance 
B 4j", what should the distance A measure? 

7. Solve ior (x) : 




5.462" 
9. Solve for (x) : 





A-? 


U B = 4f > 





3.250" 



=Q02? 



10. What size circle will just touch the corners of a rectangle 2" 
X3"? 

11. What will be the volume of a wedge shaped piece of steel, as 
shown in problem No. 7, if the width is 2"? 

12. What will be the greatest length rod that can be placed inside 
of a box 2 ft. wide X 3 ft. high X 4 ft. long? 

13. What length piece of sheet metal which is 2 ft. wide can be 
placed inside of a box 2 ft. wide X 3 ft. high X 4 ft. long? 

14. What length of wire will be required to wind around a bar 18" 
long, 2" in diameter with a 13" lead? 



Cube Root 

The Cube or third power of a number is obtained if the 
number itself is repeated as a factor three times, thus the 
cube of 4 is 4 X 4 X 4 = 64 and is written 4 3 , 

If a number is repeated 4 or 5 times, the product is the 
4th or 5th power, thus 5 4 = 5X5X5X5 = 625. 

The Cube Root of a given number is that number which, 
when repeated as a factor 3 times, will give a product equal 
to the given number, thus the cube root of 125 written V125 
equals 5, as 5 X 5 X 5 = 125. 
4 



3© INDUSTRIAL MATHEMATICS 

The 4th, 5th, etc. root of a given number are those numbers 
which, when repeated as factors 4, 5, etc. times, will give as a 
product the given number, thus V625 equals 5, as 5 X 5 
X 5 X 5 = 625. 

Example: Extract the A/33076161. 

3X3X3= 27 
3 2 X 300 = 2700 



32 (Ans.) 



(A) 3 2 X 300 X 2 = 5400 


33 , 076 , i6i 


(B) 3 X 30 X 2 2 = 360 


27 


(C) 2 3 = 8 


60 76 


5768 


57 68 




3 08 161 


« 


308 161 


(A') 32 2 X 300 = 307200 




(B') 32 2 X 300 X 1 = 307200 




(C) 32 X 30 X i 2 = 960 




i 3 = 1 





308161 

Rule. — To extract the cube root (assuming that the cube 
root of 33,076,161. is to be found) beginning at the unit 
figure or decimal point, point off the numbers in periods of 
3 figures each. 

Find the greatest whole number, the cube of which does 
not exceed the left hand period, and write this down as the 
first figure in the root (3). 

Subtract the cube of 3 from the left hand period and bring 
down the next period of 3 figures and annex it to the re- 
mainder. 

Multiply the square of the figure (3) in the root by the 
constant 300 (3 2 X 300) = 2700, and find out how many 
times this number is contained in the number 6076. 



THE CIRCLE 39 

This gives us a trial figure for the second figure of the root. 

Next subtract from 6076 the sum of the following products: 

1st — The square of the figure or figures already obtained 
in the root, except the last one, multiplied by 300 and this 
product multiplied by the figure just obtained in the root: 
(-4) 3 2 X 300 X 2 = 5400. 

2d — The figure or figures already obtained in the root, 
except the last one, multiplied by 30, and this product 
multiplied by the square of the last figure obtained in the 
root: (B) 3 X 30 X 2 2 = 360. 

3d — The cube of the last figure obtained: ( C) 2 3 = 8. If 
the sum of these various products (A) 5400, (B) 360, ( C) 
8 = 5768, is larger than 6076, it indicates that the trial 
figure is too large, and a figure one unit smaller should be 
used. 

After having subtracted 5768 from 6076 move down the 
next period of 3 figures and annex it to the remainder, and 
proceed as before. 

EXERCISES 

1. Find the third power of 321. 

2. Find the ninth power of 5. 

3. Find the fifth power of 11. 

4. Find the cube root of 24389. 

5. Find the cube root of 997002999. 

6. Find the cube root of 99700.2999 

7. Find the cube root of 0.4219. 

8. Find the cube root of 512000. 

The Circle 

A Circle is a plane figure bounded by a line 
every point of which is an equal distance from 
the center point. 

The Diameter of a circle is the distance in a 
straight line through the center from one edge 
to the other. 








4-0 INDUSTRIAL MATHEMATICS 

The Radius is one-half of the diameter or the 

distance from the center point of the circle to 

any point on the circumference measure in a 

straight line. ^^ 

A Radian is a part of a circle of such 

magnitude that the length of the arc is ,\ /%\ 

equal to the radius. 

The included angle of a radian is equal \Z x |_jL 

to 57° 17' 44.8". I*— -*' 

The Circumference of a circle is the distance 
around and is found by multiplying the diameter 
by 3.141601 roughly 3^-. The figure 3.1416 is 
sometimes called Pi represented by the charac- 
ter 7T. Example: Find the circumference of a 
circle 3" in diameter. 

Rule. — Circumference =D X ir = 3" X 3.1416 
= 9.4248" (Ans.). 

The Area of a circle can be found by mul- 
tiplying the diameter squared by 0.7854 or (J7r). 
Rule.— Area = D 2 X 0.7854. 
Example: Find the area of a circle 3" in diameter. 

A = D 2 X 0.7854 = A = 3 2 X 0.7854 

= A = 9 X 0.7854 = 7.0686 sq. in. (Ans.). 

/* ~^\ An Arc of a circle is part of its cir- 
cumference. 
A Chord is a straight line joining two points 
on the circumference. 

A Tangent of a circle is a straight line which 
touches the circumference at one point only. 

A Sector of a circle is the space between an 
arc and two radii drawn to its ex- 
tremities. 
To find the area of a sector of a circle: 







THE CIRCLE 



41 



Rule I. — Multiply the length of the arc by J the radius 

R 
= A = L X-. 
2 

Angle of sector in degrees 
Rule II. — Xtt X radius squared 



360 



A. 

360 



tR 2 . 



Example I: Find the area of a sector of a circle which 
has an arc length of 4" and a 3" radius. 

Rule I.— A = LX- = 4X-=4Xi| = 6 sq. in. (Ans.). 
2 2 

Example II: Find the area of a 6o° sector which has a 

radius of 5". 

6 60 

Rule U.—A = X tR 2 = X 3.1416 X 5 2 

360 360 

60 

= X 3.1416 X 25 = 13.09 sq. in. (Ans.). 

360 

A Segment of a circle is the space between an 
arc and its chord. To find the area of a segment 
of a circle: 

Rule I.— A = HR(l - c) + he]. 

Example I : Find the area of the following segment. 





Rule.— A = HR(l - c) + he] = |[i|(2.356 - 2.121) 

+ .440 X 2. 121] = IC1K0.235) + 0.933] 

= iCo-352 + 0.933] = £[1-285] = 0.642 sq. in. (Ans.). 

Rule II. — The area of a segment may be found by sub- 



42 



INDUSTRIAL MATHEMATICS 



tracting from the area of the inclosed sector the area of the 
inclosed triangle. 

Example II: Find the area of the segment 





of the following figure. 

The area of a i" circle = 3.1416 sq. in. 

90 = I of a circle. 

Therefore J of 3.1416 sq. in. = 0.7854 sq. in. = area of 

1 X 1 

inclosed sector. The area of the triangle = = 0.500 

2 

sq. in. 

0.7854 sq. in. — 0.500 sq. in. = 0.2854 S Q- * n - ~ are a of 

segment. 

A line drawn at right angles to a diameter 

(of any circle) will be a mean proportional to 

the two parts of the diameter divided, i.e., the 

length A : B : : B : C. 

Example: In the following figure A and 

B given as 2" and 4" respectively to find 

C. 

A: B :: B : C = 2 :^::4:8. 8 (Ans.). 

EXERCISES 

1. What area will a sector of a circle whose diameter is 2j" have if 
the chord is if" long? 

2. Two round shafts of equal diameter will just fit side by side into 
a 4!" pipe. What is the radius of the shafts? 

3. If a locomotive drive wheel is 6 ft. in diameter, how long will it 
take to travel \ mile, running at 200 revolutions per minute? 

4. If a lathe job is 4" in diameter and is turning three revolutions 
per minute, how far would a point on the circumference travel in a 
circular direction in one minute's time? 

5. If one pint of paint will cover 20 square feet of surface, how 
much paint would be required to paint a circular clock dial 12 ft. in 
diameter? 

6. Find the number of gallons of water held by a cylindrical tank 
4 feet in diameter 3 feet high. 




MENSURATION AND GEOMETRY 43 

7. If a piece \" thick is cut from the face of a sphere ^^~Tpt\ 
and the diameter of the section cut is 3", what is the di- Cl__f__3 
ameter of the sphere? L_ .J 

8. Find the area of a 15^" circle, — also circumference. [~ "A 

9. If a ball 10" in diameter be flattened on one side r ? ~ n 
to reduce the diameter to 9", what will be the area of the \ J 
flat face? ^ 

10. If a 4 ft. car wheel has a flat spot on its circum- 
ference 6" long how far will the axle drop when this spot reaches the 
rail? 

11. If boiler pressure is 150 lbs., what force will be exerted upon a 
piston 10" in diameter? 

12. Find the area of a sector of a circle, the angle being 42 deg. and 
the radius 15 inches. 

13. Find the area of a sector of a circle which has a 30" arc and a 
radius of 4 ft. 

14. Find the area of a segment which has a 2" radius, 3.1416" arc, 
2.828" chord and a rise of 0.586 inches. 

15. Find the area of a circular segment which has f " radius and the 
included sector angle is 90 . 

16. Find the area of a circular segment which has a 1" radius and 
the included angle is 6o°. 

Mensuration and Geometry 

A Point has position but not magnitude. 

A Line has length, but neither breadth nor thickness. 

A Surface has length, breadth but not thickness. 

A Solid has length, breadth and thickness. 

A Plane is a surface which is straight in every direction, 
that is, one which is perfectly flat. 

Parallel Lines are straight lines lying in the same plane 
that are everywhere equally distant from each other. Circu- 
lar lines which answer to this condition are said to be con- 
centric. 

An Angle is the difference in the direction of two lines. 
If the lines meet, that point of meeting is called the Vertex 
of the angle, and the lines are called its sides. 



44 



INDUSTRIAL MATHEMATICS 



A Right Angle is a 90 degree angle — (a-c-d). Fig. I. 

An Obtuse Angle is one which is greater than a right 
angle — (a-c-e). 

An Acute Angle is one which is less than a right angle — 
(e-c-b). 

d 



c 

Fig. I 



The Complement of an Angle is a right angle or 90 degrees, 
less the given angle. Thus, b-c-e is the complement of d—c—e. 
The Supplement of an Angle is two 
right angles or 180 degrees, less the 
given angle. Thus b-c-e is the supple- 
ment of a-c-e. 

The center of a circle that just in- 
cludes an equilateral triangle is at a 
point J of the altitude of the triangle 
from the base (Fig. II). 
A Polygon is a plain surface bounded by straight lines only. 
Triangles, rectangles, pentagons, hexagons, heptagons, octa- 




FlG. II 



gons, decagons, dodecagons, etc., are all polygons. The 
areas of irregular shaped surfaces can be found by dividing 
them into triangles, rectangles, etc. 



AEZJOOOO 



MENSURATION AND GEOMETRY 



45 



— -*••— -•* 



The Area of a Rectangle is equal to the 
length times the breadths (units being simi- §~ 
lar). Example: Find the area of the follow- L. 
ing figure: 2 X 4 = 8 sq. in. (Ans.). J 

, — , , — 1 - L The Area of a Parallelogram is 

equal to the altitude times the base. 
Example: Find the area of the fol- 
lowing figure: 3X6= 18 sq. in. 
(Ans,). 
The Area of an Isosceles, Equilateral, or Right Angle 
Triangle is equal to the length times one-half the breadth 
or one-half of the length times the breadth. Examples: Find 
the area of the following figures: 





■4'— —~ )p-0.8te'' 

4X1"= 4sq. in. (Ans.). 0.866' 

itt — 




X 



0-433 sq. in. (Ans.). 



4" Xf=2 sq. in. (Ans.). 

The Area of Any Shaped Triangle can be found by sub- 
tracting from one-half the sum of the three sides each side 
severally, then extracting the square 
root of the product of the three re- 
mainders and one-half the sum of the 
three sides. Example: Find the area 
of the following figure : 4+3+5 
= 12 = sum of three sides; 12-7-2=6= one-half the sum 
of the three sides; 6 — 4 = 2; 6 — 3 = 3; 6 — 5 = 1. 
Then V2X3X1 X 6 = V36 = 6 (Ans.). 

Two angles are equal when their intercepting lins aree 
parallel or are at right angles. Example: 




4 6 



INDUSTRIAL MATHEMATICS 



The sum of the three angles in any triangle always equa 
180 . Examples: 





45° +90° +45° = i8o° 



40 + 120° 4- 20° = i8o< 



The Volume of a Cube, Prism or Cylinder is equal to the 
area of the base times the height. 





The Volume of a Sphere = ^D 3 X it. Example: Find 

the volume of a 3" sphere. 

Rule.— V = f£> 3 X 7T = J3 3 X 3-1416 = J27 X 3-1416 

= ^84. 81 = 14.13 cu. in. (Ans.). 

The Convexed Area of a Sphere = D 2 X t. 

Example: Find the surface area of a sphere 3" 

in diameter. 

Rule.— A = D* X 7T = 3 2 X 3.1416 = 9 X 3-1416 

= 28.27 S Q- m - (Ans.). 

The Volume of a Cone = area of base X i of the altitude. 

Example: Find the volume of the following 

figure. 

H 
Rule.— V = A X - . 
3 

A = area. 

H = height or altitude. 

H 9 

V = A X - = 7.07 X - = 7.07 X 3 = 2I - 21 

3 3 

cu. in. (Ans.). 




MENSURATION AND GEOMETRY 



47 



To find the Total Surface Area of a Cone : 

H 
Rule.— Surface Area = C X — + A. 

2 

C = circumference of base. 
H = slant height. 
A = area of base. 





Example: Find the total surface area of the above cone: 

H 6 

Rule.— Surface Area =CX~+i=3X 3.1416 X - 

2 2 

+ 3 2 X 0.7854 = 9.42 X 3 + 7-07 = 28.26 + 7.07 

= 35-33 sq. in. (Ans.). 

To find the Volume of a Frustum of a Cone : 

Rule.— V = o.26iSH(D 2 + Dd + d 2 ). 
Example: Find the volume of the following 
frustum: 

Rule.— V = o.26i8H(D 2 + Dd + d 2 ) 
= 0.2618 X 5(4 2 + 4 X 3 + 3 2 ) 
= 1.309(16 + 12 + 9) 
= 1.309 X 37 = 48436 cu. in. (Ans.). 



To find the Volume of a Pyramid : 

H 
Rule.— V = — X A. 

3 
H = height. 
A = area of base. 

Example: Find the volume of the following 
pyramid: 

H 6 

Rule.— V = - X A =-X9 = 2X9=i8 

3 3 
cu. in. (Ans.J. 

To find the Area of an Ellipse : 
Rule.— A = ir X a Xb.> 




4 8 



INDUSTRIAL MATHEMATICS 





Example: Find the area of the following ellipse: 

Rule.— A = 7T X a X b 

= 3.1416 X 3 X 2 
= 18.85 sc l- i n - (Ans.). 
To find the Circumference of an 
Ellipse : 

Rule.— Cir. = tt ^ 2 {a + b 2 ). 
Example: Find the circumference of the above ellipse: 
Rule.— Cir. = tt ^ 2 (a 2 + b 2 ) =3.1416 V2(3 2 + 2 2 ) 

= 3.1416 -^2(9 + 4) = 3-1416 V2 X 13 = 3.1416 V26 

= 3.1416 X 5.099 = 16.019 m - (Ans.). 

To find the Area of a Hexagon : 

Rule.— A = D 2 X 0.866. 
Example: Find the area of the following hex- 
agon: 

Rule.— A = D 2 X 0.866 = 2 2 X 0.866 

= 4 X 0.866 = 3.464 sq. in. (Ans.). 

The volume of any irregular shaped object may be found 
by the displacement method. For example: Place an object 
in a vessel 10" long and 6" wide and pour in sufficient water 
to completely submerge the object. For convenience let us 
imagine that the water is 7J" deep; then by taking the 
object out of the water, the water will lower in level an 
amount equal to the volume of the object. Thus if the 
water was 7§" deep at first and after the object is taken out 
it is measured and found to be 6\" deep or a decrease of if". 
Thus the total displacement = 6 X 10 X ij = 75 cu. in. 
or the volume of the object submerged in cu. in. 

The following table gives the ratio of the length of side 
to the diameter of various polygons. 

Example: What must the points of a pair of dividers be 
set to, to space off a 2" circle into 5 equal parts? 

Opposite figure 5 is found the figure 0.588 or the distance 



MENSURATION AND GEOMETRY 



49 



for one inch; thus for 2" the dividers would be set to 2 X 
0.588" or 1. 176" (Ans.). 

Example: What is the length of one side of an octagon 
inscribed in a 3" circle? 

Opposite division No. 8 is found the figure 0.383 or the 
length of side for a one inch circle divided into 8 parts. 
Thus for a 3" circle the length of side is equal to 3 X 0.383" 
or 1. 149" (Ans.). 



No. of 


Degrees 


Length of Side 


No. of 


Degrees 


Length of Side 


Divisions 


of Arc 


Diameter = i 


Divisions 


of Arc 


Diameter = 1 


3 


120 


O.866 


12 


30 


O.259 


4 


90 


O.707 


13 


27 41' 


O.239 


5 


72 


O.588 


14 


25° 42' 


0.222 


6 


60 


O.500 


15 


24 


0.208 


7 


51° 25' 


0.434 


16 


22° 30' 


0.195 


8 


45 


O.383 


17 


21° II' 


O.I84 


9 


40 


O.342 


18 


20 


0.174 


10 


36 


O.309 


19 


i8° 57' 


O.164 


11 


32° 43' 


O.282 


20 


18 


0.156 



EXERCISES 

A. Draw free hand, a surface. 

B. Draw free hand, a polygon. 

C. Draw free hand, a hexagon. 

D. Draw free hand, an octagon. 

1. Find the area of the following figure: 

2. Find the area in sq. ft. of a lot 40' X 
120'. 

3. Find cost at 12 cents per sq. ft. for 
cementing a garage floor 20' X 20'. 

4. What will it cost to shingle a church 

steeple which is 12 ft. square at the base and has a slant height (meas- 
ured on the corner) of 30 ft., providing one bundle of shingles cost 75 
cents, and will cover 25 sq. ft., and full bundles must be paid for? 

5. Find the area of the following: 







7" ^ 



50 INDUSTRIAL MATHEMATICS 

6. Find the area of the following: 




8. If the center angle on a bevel gear is 28 deg., what angle would 
the back end of the tooth make with the end of the hub? 

9. What is the capacity in gallons of a cylindrical tank 3' in diam. 
and 8' high? 

10. How many cubes of iron 2" square will go into a box 3' X2' 
X i|'? 

11. Find the volume of a triangular prism 1" face, 3" high. 

12. Find the convex area and volume of a sphere 3" in diameter. 

13. Find the total surface area and volume of a cone 12" in diameter 
and 15" slant height. 

14. How many cubic yards of earth will be removed from a round 
well 4 ft. in diameter and 40 ft. deep? 

15. If cast iron weighs 0.26 lbs. per cubic inch, what will a 3" diameter 
round bar weigh 24" long? 

16. What will be the cost of plating both sides of a sheet of iron 24'' 
in diameter at 30 cents per sq. ft.? 

17. Find the area and circumference of an ellipse with major axis 
of 18" and minor axis of 8". 

18. Find the volume of a frustrum of a cone 6" high, small diameter 
2" and large diameter 4". 

19. What is the volume in cubic inches of a casting which when 
submerged in a vessel of liquid 6" in diameter, raises the level of the 
liquid 2§"? 

20. In a vessel 6" wide and 8" long are placed 5 castings. What is 
the volume of each casting in cu. in. if the level of the liquid was raised 
from 31" deep to 6f"? 

21. What should a pair of dividers be set to, if it is desired to space a 
circle 3" in diameter into 7 equal parts? 



REVIEW EXERCISES 



51 



22. What is the length of one side of a pentagon which is inscribed 
in a 4" circle? 

23. What is the perifery length of an octagon inscribed in a 3" circle? 

24. What should the dividers be set to, in spacing off an 8" circle 
into 15 equal parts on the circumference? 



REVIEW EXERCISES 

1. How many gallons of water will a tank hold which is 10 ft. long, 
6 ft. wide and 4 ft. deep? (231 cu. in. = 1 gal.) 

2. How many military paces will be required to travel 6 furlongs? 

3. How many feet are there in half a league? 

4. At 20 cents per yard for chain, what will be the cost of 4.0 fathoms? 

5. If stone weighs 175 lbs. per cubic foot, what will 5 perch weigh? 

6. How many bushels of wheat can be placed in a bin 6 ft. deep 
8 ft. wide and 10 ft. long? 

7. What horsepower is lost per hour, if 40720 B.t.u. are lost through 
the exhaust steam of a steam engine in a 10 hr. run? 

8. How many feet of lumber are there in 10 pieces of 2" X 4", 
16 ft. long? 

9. If a gas engine has a bore of 90 millimeters, what is its bore in 
inches? 

10. What will a tank of water weigh which is 3 ft. deep, 4 ft. wide 
and 6 ft. long? 

11. What will be the weight in grains of a if carat diamond? 

12. What horsepower is represented by 3,000 k.w. generator? 

13. What will a cast iron block 3" X 4" X 8" with a 1" square 
hole cored lengthwise weigh? 

14. If 1 pound of carbon burned to CO2 produces 14,600 B.t.u., how 
much energy in foot pounds should be developed by 
burning 10 pounds of carbon? 

15. What is the difference in weight between 
bars of aluminum and steel 1" X 2" X 12"? 

16. What will be the height over all of a pyramid 
composed of 4 steel balls 1" in diameter? 




PART II 

Formulas and Algebraical Expressions 

A Formula is a rule for a calculation expressed by using 
letters and signs instead of writing out the rule in words. 
The letters used (algebraical expressions) simply stand in 
place of the figures which are to be substituted when solving 
problems. 

Any formula may be transposed, that is, the value of any 
letter in the formula may be found in terms of the other 
letters by following certain set rules. Example: b = a X c 
then a = b/c, c = b/a. 

Any quantity may be moved from one side of the equality 
sign to the other by changing its sign, thus a — b = c then 
a = c + b. 

When several numbers or quantities in a formula are con- 
nected with signs indicating that additions, subtractions, 
multiplications or divisions are to be performed, the multi- 
plications should be carried out before any of the other 
operations, and division also preceeds additions or sub- 
tractions, as 18 -J- 6 + 15 X 3 — 2 = 3 -f- 45 — 2 = 46. 

When it is desired that certain additions and subtractions 
should preceed multiplications or divisions, use is made of 
vinculum , parentheses ( ), brace { }, and brackets [ ]. 

These indicate that the calculations included in the ( ) 

f } C ] shall be carried out complete by itself, before the re- 
maining calculations are commenced. If a (){} or [] 
is placed inside of one another, the one inside is first calcu- 
lated. The order of use is as follows £ { (" ) } ]. Exmple : 
2 + [10 X 6(8 + 4^2) - 4] X 2 = 2 + [10 X 6 X 10 -4] 
X 2 = 2 + [600 — 4] X 2 = 2 + 596 X 2 = 2 + 1 192 
= 1 194 (Ans.). 

52 



FORMULAS AND ALGEBRAICAL EXPRESSIONS 53 

The line or bar between the numerator and the denominator 
of any fraction is to be considered as a division sign, thus: 

12 + 16 + 22 50 

= — - 50 -r- 10 = 5 (Ans.). 

10 10 

In formulas the multiplication sign (X) is often omitted 
between symbols or letters, the values of which are to be 
multiplied thus a X b = ab and abc/d = (a X b X c) 4- d. 

A ( ) { J or [ ] may be removed by performing the 

operations indicated thus, a(b + c) = ab + ac. 

When a _ (){} or [] preceded by a minus sign (— ) is 
removed, all signs within must be changed, thus a — b(c — d) 

— a — be -|- bd. 

A period is sometimes used as a multiplication sign, thus 
a.b = a X b. 

A radical sign V indicates that a root is to be found. 
The small figure -\1 above the radical is called the index of 
the root. A radical is removed by extracting the root or 
raising both terms to the power of the index, thus, if V# = b 
then a = bXbXb = b 3 . 

In the Addition of algebraical quantities, all like quantities 
are placed in the same column, thus: ya + 2b + 4c + 36 

+ 5a + 3c = 

7<2 + 26 -f- 4c 
5a + 3& + 3c 
12a + 5b + jc (Ans.). 

When like quantities have the same sign, their sums are 
found by simply adding the numbers and annexing the 
common letter, thus: 5a + 26 + 7a + 36 = 

5a + 2b 
-ja + 3b 



12a + 5b (Ans.). 

When like quantities do not have the same sign, the 
5 



54 INDUSTRIAL MATHEMATICS 

positive (-+-) and negative ( — ) signs must be added separ- 
ately, and afterward the smaller group subtracted from the 
greater and the sign of the greater is prefixed and the com- 
mon letter or letters added, thus: Ja — 2b — 4c — 5a 

+ 36 + 3c = 

7a. — 2b — 4c 
- 5^ + 3b + 3c 

2a + b — c (Ans.). 

In the Subtraction of algebraical quantities, the signs of 
all terms subtracted must be changed, and then all like 
terms are added as above mentioned, thus: Subtract 5X 2 
— 2* + 6 from iox 2 + 4X — 2 = 



iox 2 + 4^ — 2 


1 ox 2 + 4X - 


- 2 


5.x 2 — 2X + 6 


— 5X 2 + 2X - 


- 6 



5x 2 + 6x - 8 (Ans.). 

In the Multiplication of algebraical quantities, the rules 
of signs must be observed, i.e., the product of two terms of 
similar signs is positive (+), thus: 2 X + 3 = 6, or— 2X 
- 3 = + 6. 

The product of two dissimilar terms is negative ( — ), thus: 
2X-3= -6, or - 2 X + 3 = — 6. 

When multiplying two simple expressions together, first 
multiply the figures and then add the exponents of the letters, 
thus: 406 X 2a 2 6 2 = \a b 
X 2a 2 6 2 

8a 3 Z> 3 (Ans.). 

Or multiplying a + 4 by a + 5 = a -\- 4. 

a + 5 



a 2 + 4a 

5a + 20 
a 2 + ga + 20 (Ans.). 



FORMULAS AND ALGEBRAICAL EXPRESSIONS 55 

Or multiplying 2a + 3& — %x by 2a — . 3& — 4# = 



2a +3& — 4* 




2a — 36 — 4X 




4a 2 -f- 6a& — 8ax 




— 6ab 


— gb 2 + I2xb 


— 8ax 


— I2xb + 165c 2 


4a 2 — i6ax 


- 9& 2 + i6x 2 



In the Division of algebraical quantities, the rules of signs 
must also be observed, i.e., the sign of any term of the 
quotient is positive (+) when the dividend and the divisor 
have like signs and negative (— ) when they have unlike 
signs. 

In division the coefficient of the quotient is equal to the 
coefficient of the dividend divided by the divisor, thus: 
20ab 4- $ab = 4 (Ans.) or 10a -f- 2a =5 (Ans.) or 30a 2 6 3 -4- 
- 6a 2 6 = - b 2 (Ans.). 

In the division of algebraical quantities, cancellation can 
be used to simplify the operation. Thus: 

2^a z b 2 x \2ab 
ioa 2 b x 2 $x 

The use of formulas can possibly be best explained by a 
few simple problems as follows: 

(1) Find the area of a circle 2" in diameter? 

The formula for finding the area of a circle = D 2 X 0.7854. 
D = diameter of circle. 
0.7854 = constant. 

D 2 X 0.7854 = 2 2 X 0.7854 = 4 X 0.7854 

= 3.1416 sq. in. (Ans.). 

(2) What is the S.A.E. rating of a gas engine, that has 
8 cylinders with a 3" bore, running at 2000 f.p.m.? 



56 INDUSTRIAL MATHEMATICS 

D 2 X N 

The S.A.E. formula = for a piston speed of iooo 

2-5 
f.p.m. 

D = diameter of cylinder in inches. 
N = number of cylinders. 
2.5 = constant. 

D 2 X N 3 2 X8 9X8 

= = = 28.8. 

2.5 2.5 2.5 

28.8 X 2 = 57.6 h.p. (Ans,). 

(3) What is the i.h.p. of a steam engine running at 125 
r.p.m., with a 10" bore and a 12" stroke, with a m.e.p. of 
50 lbs. per sq. inch? 

T . . , . .. 2 PLAN 

Ine formula for i.h.p. = — . 

33000 

P = mean effective pressure in lbs. per sq. inch. 

L = length of stroke in feet. 

A = area of piston in sq. inches. 

N = number of revolutions per minute. 

2 = a constant which is necessary because a steam engine 

is a double acting engine, and therefore obtains 2 

power impulses per revolution. 
33000 = number of ft. lbs. per minute, equivalent to 1 h.p. 

2 PLAN 2X50X78.54X125 . 

= = 29.75 i.h.p. (Ans.). 

33000 33000 

(4) What is the centrifugal force produced by a 10 lb. 

weight fastened to the end of a cord 5 ft. long : revolving at a 

velocity of 20 ft. per second? 

WV* 
Formula F = . 

gR 

F = centrifugal force. 
W = weight in lbs. 



EXERCISES 



57 



V = velocity in ft. per second. 
g = gravity (32.16). 
R = radius. 



F = 



WV 2 _ 10 X 20 2 10 X 400 
gR 32.16 X 5 32.16 X 5 



= 24.87 lbs. (Ans.). 



(5) What h.p. will a 2" rope transmit, running at 5000 
f.p.m.? 

Formula = h.p. = D 2 X V X 0.003 X N. 

D = diameter of rope in inches. 

V = velocity in f.p.m. 

0.003 = constant. 

N = number of ropes. 

h.p. = D 2 X V X 0.003 X N = 2 2 X 5000 X 0.003 X 1 

= 4 X 5000 X 0.003 X 1 = 60 h.p. (Ans.). 

If h.p. = D 2 X V X 0.003 X N, then the formula to find 
the diameter of the rope required can be obtained by sub- 
stitution, which will become — 



D = 



W 



h.p. 



X 0.003 X N 



= D=y]- 



60 



5000 X 0.003 X 1 
= D = V4 = 2" (Ans.). 

To find the number of ropes required, the formula by sub- 
stitution will be — 



N = 



h.p. 



D 2 X V X 0.003 



= N = 



60 



4 X 5000 X 0.003 

EXERCISES 

1. (a) Add 6a + 56 — 3a — 4& + 6a — 3& + z. 
(6) Add a+2&+c+a-&+a+3&+2c. 

2. (a) Subtract 3a 2 - 2& + 8c from 5a 2 - 4b + 6c. 
(6) Subtract i2xyH 3 from 8>xy 2 z 3 . 

3. (a) Multiply a + 6 by a + 4. 
(6) Multiply x + y by x — y. 



= I (Ans.). 



58 INDUSTRIAL MATHEMATICS 

4. (a) Divide I2a& 2 by 6ab. 

(b) Divide — 20* 3 y 2 by — 5# 2 y. 
a \p. + -,2 

5. Find the value of — when a = 3, & = 6, c = 2, y = 5. 

yb — a 2 — c 

6. Find the value of ■ ™r when x = 10, y = 5, 

32 - x — y z 2 - y 2 

2 = 7- 

T- J ..u 1 r /3ft + X fab + 7* — 12 

7. Find the value of \ ; \ : when a = io, 

\ * — b \ x — b 

b = 3. * = 7- 

8. What is the true length of one coil of a spring which is 2" in 
diameter and a |" lead? 

Formula for length of one coil = ^(27rr) 2 + L 2 . 

r = radius of mean diameter in inches. 
L = lead. 

9. What is the safe load in tons of a 6" X 6" sq. cast iron column 
10 ft. long? 

The formula for cast iron columns in tons = — 



1 "1 To 

1 000a 2 



a = sectional area in sq. inches. 
L — length of column in inches. 
d = width of sides of column in inches. 

10. What is the b.h.p. developed by a steam engine with a Prony 
brake, if the length of brake arm is 6 ft., the weight registered on the 
scale is 100 lbs. and the engine is running 120 r.p.m.? 

, , ~ , , , 2TTLWN 

Formula for Prony brake h.p. = . 

33000 

L = length of brake arm in ft. 
W = weight or pressure at end of arm in lbs. 
N = number of revolutions per minute. 
33000 = constant. 

11. The formula for the horsepower transmitted by belting is as 
follows : 

SVW 

h.p. = , 

33000 

in which h.p. = horsepower transmitted, 

5 = working stress of belt per inch of width in lbs., 

V = velocity of belt in ft. per min., 
W = width of belt in inches, 
find the values of S, V and W. 



EXERCISES 59 

12. The formula for the horsepower of a steam engine is 

2PLAN 

l.h.p. = , 

33000 

in which i.h.p. = indicated horsepower, 

P = mean effective pressure in lbs. per sq. in., 

L = length of stroke in feet, 

A = area of piston in sq. in., 

N = number of revolutions per min., 

find the values of P, L, A and N. 

13. In the formula A = ir(ab — cd) find the values of a, b, c and d. 



14. Simplify 




= 2 J U - 


-uix +z(t -m +~ 


15. In the formula 






w= A{V+S) 



)}] 



p 

find the values of A, V, S and P. 

16. In the formula for the space passed through by a falling body, 
starting from rest, 5 = \gP, in which 

S = space body falls in feet, 

g = acceleration of gravity in ft. per sec. per sec, 

t = time of fall in seconds, 

find the value of t and g. 

17. In the formula for the space passed through by a falling body 
which starts with an initial velocity 

2g 

in which 5 = space body falls in feet, 
g = acceleration of gravity, 
V = final velocity in ft. per sec, 
v = initial velocity in ft. per sec, 

find the values of V, v and g. 

18. In the formula for triangles H 2 = A 2 + B 2 , in which 

H = hypotenuse, 

A = altitude, 

B - base, 
find H, A and B. 



60 INDUSTRIAL MATHEMATICS 

19. In the formula 

z ~ s - 

find the values of M, N and S. 

20. In the formula 



D = 



N +M 
find the values of H, N and M. . 

Progression 

Progression means a progressive increase or decrease of a 
series of numbers. 

There are two kinds of progression in general use — arith- 
metical and geometrical. 

Arithmetical Progression in a series of numbers is a pro- 
gressive increase or decrease in each term by adding or sub- 
tracting a constant amount called the Common Difference. 

Thus the figures 1-3-5-7-9-11-13 etc., is an increase 
progression with a common difference of 2. 

While 1 5-12-9-6-3-0 is a decreasing progression with a 
common difference of 3. 

In arithmetical progression the following terms and 
formulas are generally used. 

A = first term, 

L = last term, 

N = number of terms, 

D = common difference, 

S = sum of all terms. 

Formulas 

L = A + (N - i)A _ L - A 

A = L - (N - i)D, " N - 1 ' 

L - A A + L 

N = i+ — — , S = — — X(L + D- A). 

D 2D 

Example: Find the last term in an arithmetical progression 
of 5 figures, if the first term is 2 and the difference is 3. 



PROGRESSION 6 1 

Formula: L = A + (N - i)D = 2 + (5 - 1)3 

= 2+4X3 = 2 + 12 = 14 (Ans.). 

Example: Find the first term in an arithmetical progression 
of a series of 5 numbers, if the difference is 2 and the last 
number is 11. 

Formula: A = L - (N - i)D = 11 - (5 - 1)2 

= 11-4X2 = 11-8 = 3 (Ans.). 

Example: Find the number of terms in an arithmetical 
progression, if the first term is 2, the last is 20 and the 
difference is 3. 

L — A 20 — 2 18 

Formula: N — 1 + = 1 + = 1 + — 

D 3 3 

= 1+6 = 7 (Ans.). 

Example: Find the common difference in the following 
arithmetical progression, first number is 3, the last 21 in a 
series of 6 numbers. 

L - A 21-3 18 

Formula: D = = = — = 3 (Ans.). 

N — 1 7 — 1 6 

Example: Find the sum of the arithmetical progression, 
if the first term is 3 and the last is 28, if the difference is 5. 

A + L 3 + 28 

Formula: 5 = X (L + D - A) = 

2D 2X5 

X (28 + 5 ~ 3) = — X 30 = 3-1 X 30 = 93 (Ans.). 
10 

Geometrical Progression in a series of numbers is a pro- 
gressive increase or decrease in each term obtained by 
multiplying or dividing the preceding term by a constant 
called the Ratio. 

Thus 1-2-4-8-16-32 etc., is an increasing geometrical 
progression with a ratio of 2 and 81-27-9-3-1 is a decreasing 
geometrical progression with a ratio of 3. 



62 INDUSTRIAL MATHEMATICS 



In geometrical progression the following 


terms and formulas 


are generally used. 






A = first term, 






L = last term, 






N = number of terms, 




R = ratio, 






.S = sum of all terms. 




Formulas 






L 


L 


= AR N ~\ 


log L — log A 

N= — — + i, 

log-R 


R 


S - A 
" S - h 


LR- A 

s = , 







R 

Example: Find the first number of a geometrical progres- 
sion of five terms, if the last number is 243 and the ratio is 3. 

it 1 a L 2 43 2 43 243 

formula: A = — $—: = = = = 3 (Ans.). 

R 3 5_1 3 4 81 

Example: Find the last term in a geometrical progression 
of 5 figures, if the first term is 1 and the ratio is 2. 

Formula: L = AR^ 1 = 1 X 2 5 " 1 = 1 X 2 4 

= 1 X 16 = 16 (Ans.). 

Example: Find the number of terms in a geometrical 
progression, if the first term is 2, the last is 486, if the ratio is 3. 

^ . AT log L - log A log 243 - log 3 

Formula: N = + 1 = + 1 

log R log 3 

2.385 - 0.477 1-908 . , ... 

= + 1 = 1-1=44-1=5 (Ans.). 

0.477 0.477 

Example: Find the ratio in the following geometrical 
progression: 1-1. 3-1. 69-2. 19-2. 85. 



Formula: R 



EXERCISES 63 

S - A 



1 + 1.3 + 1.69 + 2.19 + 2.85 - 1 8.03 

= = = 1.3 (Ans.). 

1 + 1.3 + 1.69 + 2.19 + 2.85 - 2.85 6.18 

Example: Find the sum of a geometrical progression, if 
the first term is 2, the last iof , with a ratio of \\. 

LR- A iof X i§ - 2 

formula: S = = 

R - 1 i| — 1 

= = — — = 26g (Ans.). 

2 2 

EXERCISES 

1. Find the first term in an arithmetical progression in a series of 8 
numbers if the difference is 3 and the last number is 25. 

2. Find the last term in an arithmetical progression of 6 figures, if 
the first term is 3 and the difference is 2. 

3. Find the sum of the arithmetical progression, if the first term 
is 4, the last 16 and the difference is 3. 

4. Find the number of terms in an arithmetical progression, if the 
first term is 6, the last 27 and the difference is 3. 

5. What is the common difference used in a six step cone pulley, 
if the small pulley measures 3" in diameter and the largest one 13"? 

6. What is the last term in a geometrical progression of six numbers, 
if the first term is 1 and the ratio is 3? 

7. Find the number of terms in a geometrical progression, if the 
first term is 4, the last 256 and the ratio is 4. 

8. Find the first number of a geometrical progression of 5 figures 
if the last number is 785 and the ratio is 2.5. 

9. What is the sum of a geometrical progression, if the first term 
is 8, the last 2.621 and the ratio 0.8? 

10. Find the ratio used in a five step cone pulley having the following 
diameters: 5"-6"-7"-8"-o". 



6 4 



INDUSTRIAL MATHEMATICS 



Trigonometry 

Trigonometry is that branch of mathematics which deals 
with the determination of angles and the solution of triangles. 

Any figure bound by three straight sides is called a Triangle. 
The sides are called Base (B), Altitude (A) and Hypotenuse 
(H) (Fig. l-a), or more conveniently, Side Adjacent (SA), 
Side Opposite (SO) and Hypotenuse (hyp) (Fig. i-b). All 
triangles have three angles, and their sum is always equal 
to 1 80 degrees (Fig. l-c). 




Fig. l-c 



Two angles are equal when they contain the same number 

of degrees. Example: ^- — ^ ° . — ~* — j^ 

(An angle may be defined as the path through which a ray 
passes revolving about a fixed point called the center.) 

A Right Triangle is one which contains a right angle or a 
90 degree angle. In all right triangles having the same 
acute angle, the sides are proportional. Example: 




This fact forms the working basis of trigonometry. In 



TRIGONOMETRY 



65 



practical work the right triangle is most often encountered, 
otherwise the figure can easily be subdivided into two or 
more right triangles, as shown in the following: 




There are six possible ratios between the three sides of a 
right triangle called the trigonometric functions, and desig- 
nated sine (sin), cosine (cos) tangent (tan), cotangent (cot), 
secant (sec), and cosecant (esc). These are defined as 
follows: 

(Let X equal the angle in all cases) 



side opposite 
(1) sine X - 

hypotenuse 


or 


sin 


SO 
hyp' 


side adjacent 
(2) cosine X — 

hypotenuse 


or 


cos 


SA 
hyp' 


side opposite 
(3) tangent X - 

side adjacent 


or 


tan 


SO 

= SA' 


side adjacent 
(4) cotangent X — 

side opposite 


or 


cot 


SA 
~ SO' 


hypotenuse 
(5) secant X - 

side adjacent 


or 


sec 


~~ SA' 


hypotenuse 
(6) cosecant X — 

side opposite 


or 


CSC 


hyp 

= so' 



It will be seen that the last three functions are the recipro- 
cals or inverse of the first three, and this is the easiest way to 
remember them. 

Note. — The foregoing definitions must be mastered before 
any calculations can be made. 



66 



INDUSTRIAL MATHEMATICS 



The prefix "co" in the functions comes from "comple- 
mentary," the functions of an angle being the co-function 
of a complementary angle. 

For example, sine X = cos (90 — X), tan 30 = cot 60, 
sec 15 = esc 75, etc. 

The trigonometric functions are numerical values which 
can be determined for every angle. However, it would be 
entirely impractical to determine these values for each 
application, so complete tables have been compiled giving 
the values for all angles, and the quantities are taken direct 
from the tables, sines and tangents are the two most im- 
portant functions. 

Since most calculations can be made with sines, cosines, 
tangents and cotangents, only these are tabulated in this 
book. 

-D ■ ,->j In Fig. II, the functions 

s"' \ are illustrated graphically 

R = radius = 1, 
A = sine, 
B = cosine, 
C = tangent, 
D = cotangent, 
E = secant, 
:, F = cosecant. 




Since the radius equals I, the sides or numerators of the 
ratios are disposed so that the denominator is always the 
radius, thus permitting the denominator to be disregarded 
and only the length of the numerator considered. 

Note. — (In Fig. II the angle used is 30 degrees. Compare 
the relative length of the lines with the values given in the 
tables.) 

Thus in Fig. II, if the radius is equal to 1, and the angle 
is 30 degrees, according to the table of trigonometrical 
functions: 



TRIGONOMETRICAL FUNCTIONS 67 

A = .50000, 
B = .86603, 

C = .57735, 
D = 1.7320, 
E = 1. 1547, 
F = 2.0000. 

Sines 

The Sine of an angle is the relation of the side opposite of 

any given angle to its hypotenuse, or vise versa. 

side opposite 

Rules. — 1. Sine A = -. 

hypotenuse 

2. Side opposite = sine A X hypotenuse. 

side opposite 



3. Hypotenuse = 



Problem No. 1 : What is 



sine A 



% 



the angle in Fig. Ill, if tffl*"'''^ \\ 

the side opposite = 0.68404" ,---'' "^-if"^ ' \ 

and the hypotenuse = \\^^^ ^° i 
2.000"? Fig. Ill 



side opposite 0.68404 

Rule No. 1 : Sine A = = = 0.34202. 

hypotenuse 2 

Upon looking in the table of sines, we find that sine .34202 is 
equivalent to 20 degrees (Ans.). 

Problem No. 2: What is the length of the side opposite 
in Fig. Ill if the angle is 20 degrees and the hypotenuse is 
2.000"? 

Rule No. 2: Side opposite = sine X hypotenuse. 
Since of 20 deg. = .34202. 
.34202 X 2 = 0.68404" = side opposite (Ans.). 

Problem No. 3: What is the length of the hypotenuse in 
Fig. Ill if the side opposite is 0.68404" and the angle is 20 
degrees? 



68 INDUSTRIAL MATHEMATICS 

side opposite 

Rule No. 3: Hypotenuse = — — ; . 

sine 

Sine of 20 deg. = .34202. 

0.68404" -T- .34202 = 2.000" = hypotenuse (Ans.). 

(Note that the values of the sines of all angles range from 

zero (o) to one (1) inclusive.) 

Cosines 

The Cosine of an angle is the relation of the side adjacent 
to that of the hypotenuse, or vise versa, 
side adjacent 



Rules. — 1. Cosine 



hypotenuse 

2. Side adjacent = cosine X 



,/'' s\ hypotenuse 



-».-'' ' X I \ side adjacent 

*0°°x \ 3. Hypotenuse = ; . 

.s ^S \ cosine 

S'''s' / '% ! \ Problem No. 1: What will 

f-~ £ \ — 1- the angle be in Fig. IV if the 

\* IJZ206' -J , . • Ci . ' , ., • , 

hypotenuse is 2 ft. and the side 

adjacent = 1.73206 ft.? 

side adjacent 



Fig. IV 
Rule No. 1 : Cosine = 



hypotenuse 
1.73206 4- 2 = .86603. 
Cosine .86603 is equivalent to 30 degrees (Ans.). 

Problem No. 2: What is the length of the side adjacent in 
Fig. IV if the angle is 30 deg. and the hypotenuse is equal 
to 2 ft.? 

Rule No. 2: Side adjacent = cosine X hypotenuse. 
Cosine of 30 deg. = .86603. Hypotenuse = 2 ft. 
.'. .86603 X 2 = 1.73206 ft. (Ans.). 

Problem No. 3: What is the length of the hypotenuse in 
Fig. IV if the angle is 30 deg. and the side adjacent is equal 
to 1.73206 ft.? 



TRIGONOMETRICAL FUNCTIONS 69 

side adjacent 

Rule No. 3: Hypotenuse = ; . 

cosine 

Side adjacent = 1.73206 ft. Cosine of 30 deg. = .86603. 

.'. 1.73206 -r .86603 = 2 ft. (Ans.). 

(Note that the values of the cosines range from zero (o) to 

one (1) inclusive.) 

Tangents 

The Tangent of any angle is the relation of the side opposite 

to that of the side adjacent, or vice versa. 

side opposite 

Rules. — 1. Tangent = — . 

side adjacent 

2. Side opposite = side adjacent X tangent. 

side opposite 

3. Side adjacent = . 

tangent 

Problem No. 1 : What is the angle in Fig. V if the side 



I 

14' I c 



1 'St 



k 1375" >l 

Fig. V 

opposite is equal to 1.96347" and the side adjacent is 7.875"? 

side opposite 

Rule No. 1 : Tangent = — ; . 

side adjacent 

1.96347 ■*■ 7.875 = .24933. Upon looking in the table of 

tangents, we find that tangent .24933 is equivalent to 14 

degrees (Ans.). 

Problem No. 2 : What is the length of the side opposite in 

Fig. V if the angle is 14 degrees and the side adjacent is 

7-875"? 

Rule No. 2: Side opposite = side adjacent X tangent. 
Side adjacent = 7.875". Tangent of 14 deg. = .24933. 
7-875" X .24933 = I-96347" = side opposite (Ans.). 
6 



7o 



INDUSTRIAL MATHEMATICS 



Problem No. 3: What is the length of the side adjacent in 

Fig. V if the side opposite is 1.96347" and the angle is 14 

degrees ? 

side opposite 

Rule No. 3: Side adjacent = — . 

tangent 

Side opposite = 1.96347". Tangent of 14 deg. = .24933. 
.'. 1.96347" -f- .24933 = 7-875" = side adjacent (Ans.). 
(Note that the value of tangents range from zero (o) to 
infinity.) 

Uses of Inverse Functions 

It frequently happens that an inverse function may be 
used to advantage, for example, in Fig. VI, if it is desired to 
find the length of face (or hypotenuse) of a cone for a clutch, 
the following formula probably would be used: 

\ _ — - — ' I/O' 




Fig. VI 



Hypotenuse = 



side adjacent 



cosine 



.98481 



= 3.0462" (Ans.). 



This would be a rather long and tedious operation, and since 
the Secant is the reciprocal of the cosine, the process could 
have been shortened by multiplying the side adjacent (3") 
by the secant of 10 deg. (1.0154) which equals 3.0462" (Ans.). 




EXERCISES 
Solution of Right-angled Triangles 



71 



Sides and Angles 
Known 



Formulas for Sides and Angles to Be Found 



Sides c and a . 
Sides c and b . 
Sides a and 6 . 



b =Vc* - a 2 

a = 1/c 2 - 62 



Side c. Ang. A . 

Side c. Ang. B. 

Side a. Ang. A . 

Side a. Ang. 5. 

Side b. Ang. 4 . 

Side b, Ang. B. 



c = i/a 2 + 62 



c X sin A 
c X cos B 



sin A 



cos B 
b 

cos A 

b 
sin B 



sin A = 



c 

. ^ b 

sin B = - 

c 

a 

tan A = - 
6 

6 = c X cos A 
b = c X sin B 

b = a X cot A 

6 =aXtanB 
a = 6 X tan A 
a = 6 X cot .B 



B = 90 - A 

A = 90 - £ 
B - 90 - A 



5 - 9 o c 
A = 90 c 



B = 90 - A 
A = 90 - B 
B = 90 - A 



90' 



B 



EXERCISES 



1 . Solve for (x) : 




4. Calculating the circumference from root diameter and allowing 
5 deg. side clearance on the tool, what will be the actual cutting clearance 
on the right and left side of a square thread tool used to cut a 4" root 
diameter and two threads per inch single thread? (Fig. VIII.) 

5. At what angle from regular position will the table of universal 
milling machine be set to gash a worm wheel meshing with a worm 
having 4 threads per inch double thread and 2" P.D.? 



72 



INDUSTRIAL MATHEMATICS 



6. If in sighting over a 30 deg. triangle, your sight would strike a 
window in a tower, the base of the tower being 200 ft. away, how high 
would the window be from the base of triangle? 

7. If two public highways join at a certain point, one running at 
an angle of 32 deg. south of due east and the other 27 deg. north of east, 
how far would a man walk if he left a point 1 mile from the junction 
and traveled at an angle of 90 deg. from south road before he came to 
the north road? 

8. If in cutting a taper of 4 degrees included angle, on a piece 10" 
long, how far over would you set the tail stock on a lathe? 




Fig. VIII 




9. A machinist in boring three holes in a jig finds the distance from 
A to C is 1". The hole B is at right angles with A-C and a line 30 deg. 
from A-C will intersect B. What is the distance from B to C? 

10. Using same illustration as in problem No. 9, if the distance from 
B to C was 2", what would be the distance from A to C? 




EXERCISES 



73 



3. Solve for (x) : 




4. How many \" steel balls will be required for a bearing, allowing 
0.002" clearance between each ball if the 
inside ball race has a diameter of 1.440"? 

5. What diameter of outside ball race 
will be required on a bearing, using 15 balls 
|" in diameter, 0.002" clearance between 
balls? 

6. The guy ropes of a flag pole make an 
angle of 35 deg. with the pole. If they ex- 
tend 30 ft. from the base of the pole, how 
long are they? 

7. What included angle corresponds to a taper of 5.10" per ft.? 

8. If by elevating a transit placed 10 ft. above sea level to an angle 
of 12 deg., it strikes the top of a tower 4000 ft. away, how high is the 
tower? 

9. What will be the respective dimensions of the sides of a rectangular 
base that will just fit in a 3" circle, if the sides are in a ratio of 2 to 1? 




A= Outside Diam. of Ball Race 
B= Inside Diam. of Ball Race 





10. What is the smallest size rectangular box that can be used to 
pack seven 1" steel balls, arranged in a circle? 

11. What must be the included angle of a Rivett Dock 8-pitch V 
threading tool to cut a 60 deg. thread, if the cutter is 3" in diameter and 
is ground \" below center? 

12. From the top of a cliff 2000 ft. high, the angles of depression of 
two ships at sea are observed to be 45 deg. and 30 deg. If the line 
joining the ships point directly to the foot of the cliff, find the distance 
between the ships. 

13. Two ships leave the harbor at the same time, one sailing northeast 



74 



INDUSTRIAL MATHEMATICS 



at the rate of 8 miles per hour, and the other sailing north at the rate of 

12 miles per hour. Find the shortest distance between the ships if 

hours after starting. 

• 14. What is the spiral angle of a spring 3" in diameter and with a f" 

lead? 

15. Fig. I given c 6", B 15 deg., find A, a and b 

16. Fig. I given b 0.852", A 14 deg., find a, Bex 
and c. 

17. Fig. I given c if", b if", find B, A and a. 

18. Fig. I given c 12", A 42 deg., find b, a and 
B. 

19. In the bushing plate shown in Fig. II, calculate the distance 
between the following holes: A to B, A to C, C to D, and C to E. 

20. In Fig. II, calculate the distances U, W, X, Y, and Z. 





Fig. II. 



-6.000' 
Bushing Plate for Drill Jig 



Speeds and Feeds 

The proper peripheral speeds and longitudinal feeds are 
necessary in machine tool operations to obtain the maximum 
production without heating the cutting tools to a softening 
point. 

By Peripheral Speed is meant the distance through which 
a point on the periphery of the work or tool would pass in a 
given length of time, and is usually designated in feet per 
minute (f.p.m.). 



FEEDS AND SPEEDS 75 

The peripheral speed of a drill may be determined by 
multiplying the circumference in feet by the number of r.p.m. 

In all turning operations on a lathe, there are three very 
important questions to be considered: the cutting speed, 
the feed of the tool, and depth of cut. 

The r.p.m. for any work on a lathe can be found by multi- 
plying cutting speed by 12 and dividing the product by the 
circumference of work in inches. 

Cutting Speed is found by multiplying r.p.m. times 
circumference in inches and dividing by 12. 

The cutting speed on a milling machine = circumference of 
cutter times r.p.m. 

The peripheral speed of grinding wheels = circumference 
of wheel in feet X r.p.m. 

Approximate Formulas 

To find the revolutions per min., if a certain cutting speed 

is wanted, 

cutting speed X 12 

r.p.m. = ; 

t X diameter 

or 

3.82 X cutting speed 

r.p.m. = ; . 

diameter 

To find the cutting speed in ft. per min. if r.p.m. is given, 

7r X diameter X r.p.m. 

cutting speed = ■ — . 

12 

To find the total time to finish cut. 

length in inches 



Total time to finish cut = 



r.p.m. X feed per rev. 



7 6 



INDUSTRIAL MATHEMATICS 



Chart Showing Approximate Cutting Speeds in Feet per Minute for 
Various Machines and Materials 







High Speed 


Tool 






Steel Tools 


Steel Tools 


Material 


Machine 










Speed in Feet 


Speed in Feet 






per Minute 


per Minute 


Tool steel 


Drill press 


50-60 


20-30 




Lathe 


50-70 


25-35 




Miller 


50-60 


20-30 




Shaper 


40-50 


20-25 




Gear cutter 








Planer 


40-50 


20-50 




Screw machine 


60-70 


25-35 


Cast iron 


Drill press 


IOO-170 


40-80 




Lathe 


75-175 


40-80 




Miller 


IOO-150 


60-80 




Shaper 


80-IOO 


5O-60 




Gear cutter 


60- 80 


30-50 




Planer 


70- 90 


40-50 




Screw machine 


IOO-150 


50-70 


Machine steel .... 


Drill press 


100-120 


50-60 




Lathe 


100-150 


50-70 




Miller 


IOO-I25 


50-70 




Shaper 


60-80 


50-60 




Gear cutter 


60-80 


30-40 




Planer 


50-70 


40-50 




Screw machine 


I0O-I5O 


50-70 


Brass, bronze .... 


Drill press 


200-300 


IOO-150 




Lathe 


150-300 


70-I50 




Miller 


150-250 


80-125 




Shaper 


100-120 


60-80 




Gear cutter 


100-125 


50-60 




Planer 


90-100 


60-70 




Screw machine 


200-300 


IOO-150 


Aluminum 


Drill press 


200-300 


IDO-150 




Lathe 


200-300 


IOO-150 




Miller 


200-350 


IOO-175 




Shaper 


125-200 


80-125 




Gear cutter 


150-200 


70-IOO 




Planer 


150-200 


75-100 




Screw machine 


200-300 


100-150 



The above speeds should be increased or decreased according to tl o 
nature of the work, tool, lubricant, machine, etc. 

EXERCISES 

1. What is the peripheral speed of a 1" drill traveling no r.p.m.? 

2. In drilling a piece of brass, the spindle makes 3500 r.p.m., and 
a 1/16'' drill is used; what is the peripheral speed of the drill? 



FEEDS AND SPEEDS 77 

3. In drilling cast iron, using a 1" drill taking a 0.007" feed and 
130 r.p.m. how long would it take to drill through a piece 6\" thick? 

4. How long will it take to make a cut on a piece 10" long, in a lathe 
with a 0.020" feed running at 180 r.p.m.? 

5. Find r.p.m. if a cutting speed of 50 ft. per min. is wanted and 
the diameter of cutter is 6". 

6. Find cutting speed of a 6" diameter cutter which runs 60 r.p.m. 

7. How long would it take to shape off \" of the surface of a piece 
6" long by 4" wide, feed 0.010", cut \" deep, and it takes 2 sec. to 
complete the stroke? 

8. How many r.p.m. would a 6" diameter emery wheel on a surface 
grinder make, whose peripheral speed is 6000 ft. per min.? 

9. At what speed should a \" drill be run to have a cutting speed of 
50 ft. per minute? 

10. At what speed should a £" end mill be run to obtain a cutting 
speed of 80 ft. per minute? 

11. What is the cutting speed of a lathe tool if the work is 3" in 
diameter, running at 100 r.p.m.? 

12. What is the cutting speed of a milling cutter 8" in diameter, 
running at 60 r.p.m.? 

13. How long will it take to finish a cut 18" long with a 1" end mill, 
running at 80 ft. per min., with a 0.010" feed? 

14. How long will it take to make a cut on a shaft 14" long, using a 
feed of 0.080" on a lathe running at 120 r.p.m. 

15. What is the time required to drill a f" hole 1" deep in 1000 
brackets if a feed of 0.010" is used, and the drill runs 300 r.p.m. pro- 
viding it takes 10 seconds to load the jig? 

16. How long will it take to cut a keyway 8" long in 750 shafts, if 
the cutter runs 240 r.p.m., with a 0.018" feed, allowing 30 seconds 
time for loading fixture? 

17. What speed should a gear cutter run that is z\" in diameter to 
have a surface speed of 90 ft. per minute? 

18. What is the cutting speed of a lathe tool if the work is 2\" in 
diameter running at 185 r.p.m.? 

Cost Calculation 

In large plants, before a machine is manufactured each 
individual part is classified separately by the planning de- 
partment and a Tool Operation Index Sheet is made out 
as shown. 



78 INDUSTRIAL MATHEMATICS 

These operations are so arranged as good judgment and 
past experience shows will be the best and most satisfactory 
way of handling the particular part in view. 

This sheet gives the tool designer the sequence in which 
the various operations are to be performed and he can thus 
design his tools and fixtures accordingly. This sheet also 
gives the necessary machines required and the order in 
which they should be arranged in the department to facilitate 
production and prevent any unnecessary waste of space and 
trucking. 

When things are running smoothly in the shop a Motion 
Study Record, as shown, is taken by the time study depart- 
ment of each operation by means of a series of close observa- 
tions timed by a stop watch. The duration of this study 
should be sufficiently long enough to get fair average results 
from which bases the piece work price is determined. This 
price usually holds good as long as the method, machine, 
tools or other working conditions are not changed. 

The motion study record is followed by an Instruction 
Card which gives to the foreman, set up man and the oper- 
ator, all instructions relative to tools, jigs and fixtures 
necessary to perform an individual operation. Also giving 
time allowance predetermined by a time study, the indi- 
vidual time for each elemental operation, feeds and speeds, 
etc. The foreman should see that this card is followed out 
in detail unless a better or more efficient way is found. 

For the foreman's information a Production Routing 
Sheet is sent him from the planning department, which 
gives to him in a classified form the various operations per- 
formed in his department, with the piece price of each 
operation and other information which he should keep on 
hand for future reference. 




Automatic Screw Machine 






_.6 



ABAjng-^so^ 



6u'd8ax-2ai)i 



+uojno^s2y 



soiooy 
4Sa a ^gqn| 3 



|0Gl(3S 



+5J'j=g|DfidsoH 



^.'ii-Olooipaw 






|uaoiAo|dui3 






^9P°M+z>W 



■9bui-iaaui6u3 



35} buiscqDjrvj 



2:uou34.u;c^ 



_: ; '._;-J'.c 







>.E 


• 1 


o.-e 










co< 


c 


*-<!> 


























• 1 












OS 


K'C 










<S 


£ 


1— o 

















_E 


• R 


orE 




+■ 




00 


*" 






vT 


.s 




^ 




■*; 




E 

o 
c 

a: 


a 


X 
10 

> 
a 


m 


15 
E 


.t?E 




o u. 
1— o 


L 


o 








< 


Sj 


• 1 


o- 

o 1- 






>.F 


*J 






JC 






£-= 


• & 













^tS 


c 


1— <-> 




^ 




r ^ 


— 










igs's 


• & 






)r, 




o 


tc 3 


£ 


t— o> 


1 


-£ 






•ci : 


o 






^ g 








o 


io 1 — 


(— l^ 


u 










5 


■s !• * 


o£ 


E 








J! ! -E 


^-■l 



If 
1 



UJ 

D 
O 

I 

a 




" — 
O ^ 

o Q " 

■*- O- 

o 









a 
o 
i 
ifl 

UJ 



a 
o 

u. 


= ^J 




c 

c 

c 


o 

I. 
c 

CO 

.c 

B 

O 
(XL 


I/] 

i_ 
c 

o 

o 


1 

1 


e>> cy cy 

O Jr 

2C Cl£ 




CSn£ 

2: 














o 

o 



EE 



t/J 

o 

r 

(0 

ID 
Z 
UJ 

10 
I/O 

< 




• 


| 


o 
■o 

c 


cy 

[o 

E 
c 

C 

i- 

-j. 
u 
o 


JJ^ 


• 


| 




• 


| 




• 


% 


.2 «5 g 

x E S 


• 


| 


4JSJ 


• 




-2 of 

^ c Si 


• 


!• 


■hi 


• 


| 



> a- 



• I 



^■5 t o» S- 

1|!||4 






• I 



So|t# & 



COST CALCULATION 



79 



TOOL OPERATION INDEX SHEET 
Part Name Piston Material Cast iron Part No. 1024 



Oper. 




No. 


Name of 




Mach. 


No. 


Operation 


Req'd 


Mach. 


Size 


No. 


I 


Inspect casting for 
hard spots, cracks, 
etc. 


I 


Bench 


10 ft. 


v 


2 


Rough turn diam., 
face ends and 


2 


Potter & Johnson 
piston mach. 




409 








411 




rough groove 










3 


Anneal and season 


1 


Std. gas furnace 


No. 636 


1108 


4 
5 


Water test 


1 


Special fixture 
Garvin horizontal 




1097 


Drill and mill in- 


I 


2 spindle 


1048 




side faces of bosses 




drill press 






6 


Bore and face open 
end 


I 


Warner & Swasey 
hand screw mach. 


No. 4 


971 


7 


Center end, finish 
face end, rough 
and finish grooves 


2 


Potter & Johnson 


5A 


189 
200 


8 


Finish mach. wrist 
pin hole 


1 


Koefer special 
drill press 


Special 


991 


9 


Roll piston ring 
grooves, chamfer 
grooves and both 
ends of piston 


1 


Reed lathe 


14" 


124 


10 


Drill oil grooves 


1 


Leland & Gifford 
drill press 


1 spindle 


78i 


11 


Drill, spot face and 
tap lock screw hole 


1 


Leland & Gifford 
drill press 


2 spindle 


544 


12 


Finish grind diam. 


I 


Norton 


10 X 36 


234 


13 


Inspect 


1 


Bench 


10 ft. 


V 


14 


Grind relief 


I 


Norton 


10 X 36 


236 


IS 


Grind wrist pin hole 


2 


Bryant 


No. 18 


1102 


16 


Cut off center and 
face end 


I 


P. & W. 
shaving mach. 


6" 


337 


17 


Polish end 


1 


Speed lathe 


6" 


675 


18 


Inspect 


I 


Bench 


10 ft. 


V 


19 


Oil and place in 
cartons 


I 


Bench 


5 ft. 


V 



8o 



INDUSTRIAL MATHEMATICS 



P 

o 
o 

s ^ $ 

>* 2 s « 

Si • » 

i_ eS „ «-. -m 
K "C tJ « » 

w 09 P< O Q 
H 

O 

* 



=3 




£ 






eu 


-a 
-3 




CO 


bo 


r, 




3 


CU 


T3 


n 




a 


I- 


o 


aS 


T) 


J3 


a 


3 


>. 




CJ 


w 


u 




rt 


0) 

cu 

n 


T3 

a 

03 


09 


rrt 


)h 


T3 






fl 


■a 


ft 


aj 








lO 


£ 


CU 

e 


O 


3 






< a 



•o o -* o 



£ 3 £ 



T3 O ro 

•a *""• ■* 



CU 



O M 
£ °* 

O M 

o 



a S 



> .a t-, 

+* Sf a 



w 
o 

P « 

w 
o 



S .2 






Pi o 



S 



>> 








i/"> 


ro 


o 






ro O 


ro io 


(S 


^N 






M 


o 


00 






O -* 


q q 


E 
6 






















a§B 






iO 


ro 


lO 






00 O 


ro »o 


3 










O 


a 






Os "* 


o o 


w 






























lO 


ro 


N 






CO I> 


ro tJ- 




•ApUI 






H 


o 


o 






O "* 


o o 


a 




























»o 


00 


o 






ro O 


ro i> 


H 


•;uo3 






" 


" 


CN 

H 






M M 


*T *? 










io 


ro 


Tj" 






m r— 


ro »o 


a 


•Apuj 






M 


q 


ON 






q ^t 


q q 








U^ 


00 


CM 






ID CN 


lO O 


H 


•^uo3 






M 


M 


M 
H 






m -q 

M H 


\q t^- 

M HI 










"* 


ro 


lO 






ro t- 


ro n- 


o> 


•Apuj 






M 


O 


Os 






O rt 


o o 


s 




























-* 


t^ 


N 






U-) CM 


lO O 


H 


•;uo3 






M 


M 


M 
M 






M O 
HI HI 


NO vO 










VO 


r-0 


CN 






ro ro 


(OO 


S 


•Apuj 








q 


Ov 






q ^t 


q q 








O 


0\ 


H 






^f r* 


o \o 


H 


•}uc>3 






HI 


H 


M 






M IT) 

M M 


vq vq 










lO 


fO 


lO 






ro ^r 


ro J> 


s 


•Apuj 






M 


q 


q\ 






q tj- 


q q 








lO 


00 


ro 






vO O 


ro o 


H 


•^1103 






M 


M 


M 






h vq 


M M 




















CN 


H 




















o 


o 




;iui]q 
















o 


o 

-H 


u 






















m^.a 




Htf.T 










»-|00 






Tf 


-* 














fO 






















O 








T3 




paajl 










q 








a 

X! 


F 


)aad§ 










o 
o 






o 

o 


O 

o 


s 


up;no 










<N 






o 

HI 


o 

H 




c 
_o 

.2" 
o 

to 


0) 
CO 

"o 
o 


•a 

<D 

0) 
bfl 

a 

a! 


; o 
! m 

• .s 

: & 


T3 O 
CD <N 

S? fi 

a p 


a . o 


be 

T3 
,0) 


CU 

03 
bo 

a 

<u 

-a 
a 


' vC 


o 
o 
-u 

g 

r^2 








Q 


u 




00 


? • 2 


T) 


KJ 


• ro ** 

• « gi 

• cu a 

sal 




w|oo 




13 
a 
E 

0) 

3 


0) 

u 

cu 

ft 

(U 

u 


33 
B 

cu 

3 
O 


• a 

al K 

cy cu 


TO ^Q 

o a 

^^ 

H 


T3 a 

. -t-> TJ 


a 

CS 

OJ 
CU 

a 


^4 

o 

o 

■M 


2 If! 






Ph 




a 


a 






Q£ 


i 




6 


H 




OJ 


ro 


^ 






lOO 


I> 


1 


fc 

























COST CALCULATION 



81 







PO fO (N 


t-- m 


to 


a 
S 


*»N 


q q °. °. °. 






no ro N t> 








o o o o o 




OT 


-J3AV 










rt to fO N t*« 




V 

B 


•ApUJ 


q q q q q 


<0 




m o 0\0 ro 


H 


•;uo3 


oo oo oo On O 


H 






MM MM CS 


OS 

3 








01 

S 


ApUI 


o o o o o 








OJ 




ro oo o m h 


H 


;uoo 


r- t— oo oo 0\ 


eS 






MM MM M 


o 
H 






ro m ro O O 




ApUJ 


q q q q q 






N I> O O N 




H 


•;uoo 


r^ I— 00 00 0\ 








MM MM M 


> 








01 

a 


•ApUJ 


q q q q q 


s 




oo o ■"*• o 


H 


•luoo 


o r- t- oo oo 

MM MM M 








roo ^J-O O 




•ApUJ 


q q q q q 


> 




ro Ov ro O m 




H 


•;uo3 


t» t^ 00 oo Os 






MM MM M 


P. 
















w 


CO 


pn 










U 


mtf.a 


•Hi-* 














ms.T 




d 






-a -a 




P39J 


3 d 
0J oJ 


1 






,3 X! 


4> 
O 

P. 
CO 

o 


1 


333dg 


o c 
o o 


3 


innrQ 


o o 






— • ■ . • • - -M X 






o 








•d.3 o 






a 


o 








8 °^ 


P. 










o 




^ jjc 






















a 


a 








5f 05 a. 






s 

Q 

c 
u 

S 


fa M 

+-> C 


1- 

a. 
C 
a 


U 

+J c 


& 

C 

c 


back partir 
Dsen collet, t 
ece and plac 


d 
0) 

<J 
o 

CO 




w 


^ 2 ^ * 3 j ~ a 


OS 






« 








4) 




o 


00 OO M CN rorflOO N00 


CO 




z 






M 


H 


M M M M M M M 


m 



o\>. ^ * 

CO CO M ° ^" 



A 



<£ 



WW 

0\ 



(0 0) <o CO 

8 g .§ S 8 .s 

g SHSS S^ 

5^2 © © 1 • 



COM _ g 



4>H 



•a o C to 

• 0> 

fO S ^> cD 03 S H 
O a m > > el «-i 

•r* ri &0 ^".3 3 

4> I ? 



d g a* 

.2 .a m 

^ H &g 

XI 



01 

d « 4) 

CO 3 o 

fc S S 2* 

O O *> 



d 

o 

M 

m 

a> 



O « 



■S > * «> > 

^ ® S o w 
4) W g 5 s, 

3-5 O^ S 

oofa«j<j 



82 



INDUSTRIAL MATHEMATICS 



<u o 



3 £ 

«■§ 

CD <u 

Is 

o G 

2 as 

M Q 
•SO 



G O 

<u o jr 
PQ PQ Q 



<d £ 

6 I 
rt .53 

W) C3 



G 3 5 

— ' a» G 



o fe fc 

°££ 

<+h <U CD 

<y c Mi bo 

g "z> a o, 

"4 3 G G 



° cd o 

i- bO"-*-' 
OJ be o 
3.3 § 

OPnffl 



EKE 





o 


£i 


^°°olC X 


_ai 


o\ t- -— {r h 


^ 


f0^tf5^ w 


N • X 





CD 








G 








O 








£ 








-<-) 








G 








eU 








o 
















*C 








G 




"3 




H* 




•o 

G 

£ 

G 
O 

■d 

8 

G 




^J 




-G 








to 




O 

(72 




"5 
<G 




a 


#> 


<L) 




o 


CO 


-Q 




'•5 




O 


00 


G 


a 




t^ 


o 




to 


O 


4> 




4) 








U 








4> 
















s «. 








«M 








o >, 








oS 


jQ 






fc 


H|N 






. 


*J 






X 


-G 






>2 


.s? 






^ p< 


'5 






G P. 


£ 






+J (V 








o. 








'C A 








o ■** 








W "O 
















u. 0\ 


G 

o 






o M 






t-i 








o 


to 
U 






OS 


, 






4) H 


ttf 






>8£ 


4> 






» No. 

ndura 

-LengI 


"3 






EW 


O 


.G 




Fixtu 
Tool 
Belts 
Kind 


4> 

J* 

CO 





t_ 


IU 




N>K 


/•! ! 


+ 


1 
„0ZL ' 


"1 i ]f ~ 




,018' 
,0$8> 




<- J 









COST CALCULATION 



83 



© 

I 

CO CO 



^ a ^ 






N a> o 

6 t3° 










a 



^ a 



• a 

•0 3 -2 

as 1 

p. 



S 



u 



5 • £ 



Ills' 1 
5a * 



«» 




































a; 










JZ 


























-1 






c. 


O 






ro 







OJ 


OJ 


t» 




cd 


00 

H 


0> 

m 






6 

0) 


.3 




3 




6 







bfl 




£ 


X 


3 


£ 


1 


DQ 


3 


4_> 


toO 




co 


H|N 





CO 


HH 



o 

o 

1 1 



2 x 

(3 

« « 3 



3 

o x) 



co 



;-' O 






.3 3 
, m V 



S o 

J2 O 

p O 

O CJ 



.1 

If 



"o o 



a; .3 

w x ^ 

*3* ft 



4-> 0) 



UHcoO 



3£^ 

i« w 3 O 

ffl ™ O i 3 
Oh CO g OP4 



o3 -S O 



5 2 ft 
•2" S 

a t> ft 

wj c a 

oj o .- 

co to to 



w a _j 

O 3 -p 

0*2 3 g-o 



c* 



O^O 



3t3 



2 «? 2 

_2 55 "53 53 

CO tf tftf 



* 3 > CO 



w 




Oi 


V) 


a 


M 




<a 


CO 


a 


u 


4) 


to 


M 



84 



INDUSTRIAL MATHEMATICS 



10 C/3 



W § 
W * 

W 

H 

P 
O 



w 



3 


CO 

a 


S 


H 


3 








< 




13 


00 


Li 




<u 


u 


OS 




a 


o 

S3 





a 




cfl 


03 
CJ 


O 


^ 




CI 




rt 


00 


u 


^ 


6 




a 


H 


fc 


M 


S 


4) 

OS 


A 


Q 



•JH 




H« 










J3d UOt} 


0\ 


CO 


H 


•*t On 00 


0\ vj 


OO 00 


-onpojj 






M 












M 


00 


o 






9DUJ 


O\oo rf 


CN t- 


t— CO CO On 00 m 


O On 00 CO 


"O O 0\ O Ov 


O 


O CN 


CN O 


O cn cn O O m 


h o o o 


O M O M O 














939IJ 


OKOKOffiOOOKOffiOffiOOOEOK 




o 


o 


£ 


cn cn co 


co *t 


\T) tO 


uop 


l 


U 


O 


u u u 

M H M 


1 1 

U U 


1 1 


•;daa 


o 




fO 


^3- 


ro 


t- m o 


rf On 


0\ o 




t^ 


<M 


o 


o t^ o 


"<t O 


M CM 


•om 


Tf 


M 


CN 


M rj- m 


CN M 


rf Tt 


ainpaqos 


o 


O 
M 


o 
CN 


O O O 
CN CN CN 


O O 


o o 

CN CN 




H 


H 


H 


H H H 


H H 


H H 


J9Clp H 


- 


- 


M 


■>• M M 


H > 


M M 


JO}B 

-jadO 


H 


- 


M 


CN M H 


M N 


H H 










fc ~ 


*G 


c a 










O o 




a a 




CO 
+J 

Of 


CO 




2 "2 


"^ !? 




OJ 






c a 


o 


^ ^ 




co 
M 

d 


CO 


•a 


ft, ge 

loles 

ump 


rew h 
nd 
lowei 


o o 
•a xi 


s 

a 
O 


CJ 

*c3 

a 
e 

a 

OJ 

J 

IS 


OJ 

C 

S 
6 

"a; 


o 

00 

c 

CO 

>-> 

.£! 
■+-> 


Hand ream cam sha 
shaft and cam shaft 

Machine ream water p 
pump pads 

Face synchronizer end 


il and ream lock sc 
and face propeller e 
assemble upper and 


11 and counterbore 
oles in lower half 
11 and counterbore 
oles in upper half 
pection 




2 


rt 
£ 


to 


'u -^ .2 

Q P 


• *H * "C •« eg 
Q Q 5 


SSlIBQ 












M H >• 


JO -on 
















Tt 














00 


00 












l> 


r^- 












T3 




i_i 


t^ M 


i> 


«!f o 


*°N 


o 


"> O 00 


o ■> 


CO oo > 


•qoBp\[ 


CO 


ro Os 


t> 


O »0 




O 


0\ 












O 


o 












fO 


M 












00 


o 


H 


t^ OO vO 


00 


O l> 


•ON 


"t 


t^ 


o 


i- ■**■ o 


t- > 


00 lO --, 
lO O ^ 

M 


S!f 


o 


co 


On 


On On "* 


CO 


•ON 


M 


CN 


CO 


rt to o 


l> 00 


ON O M 


•J9dQ 












M M 



COST CALCULATION 



85 



Cost Calculations 

In estimating the cost of operating a machine shop, the 
cost of labor, material, overhead expenses, etc., must be 
considered. 

Overhead Expenses include power, heat, light, insurance, 
depreciation, supervision, etc., and all other expenses which 
are not classified under labor or material. 

The following estimates will be approximately correct for 
a shop employing 12 men, and doing tool work, such as dies, 
jigs, gages, fixtures, etc. 



Cost of Materials 

High carbon steel (tool steel) 12 cents per lb. approximately 

High speed steel 90 " 

Machine steel (low carbon steel) 5 " 

Screw stock (low carbon steel) 7 

Cold rolled steel (low carbon steel) 5 

Cast iron-castings 3 

Bronze 25 " 

Brass 17 " 

Copper 20 

Aluminum 60 

Lard oil. ... 63 " " gal. 

Machine oil. . . .' 13 " 



Cost of Labor 

Three first class toolmakers 75 cents per hour 

Three second class toolmakers 60 " " " 

Two lathe hands 50 " 

One utility man 40 " " 

One apprentice 20 " " " 



Overhead Expenses 

Light $3.00 per month 

Gas for furnace $4.00 " 

Heat $5.00 average per month 

Power Approx. 30 cents per hour for 10 h.p. 



86 INDUSTRIAL MATHEMATICS 



r 

S IT 



foreman of room .... $0.80 per hour 

Supervisions manager $1.00 " " 

clerk. . $12.00 per week. 



Insurance 



J Accident \\% of pay roll. 

' Fire $5.00 per $1000 per year. 



Rent — $25.00 for room 40' X 40' 

Depreciation on machinery — 10% 

Interest on money — 6% 

Cost of investment: 

Two lathes — 14" and 18" $1600 

One universal milling machine 1300 

Hand milling machine 250 

Shaper 350 

Speed lathe 75 

Gas furnace 75 

Bath grinder 800 

Two drill presses 250 

Miscellaneous tools: drills, reamers, counterbores, etc 200 



Total . $4900 

It would require an investment of $5000 for equipment; 
the payroll for 90 days would be $3000, cost of materials and 
overhead expenses $2000, also a surplus of $2000 to take 
care of bad debts and any emergencies, making a required 
capital of $12,000. 

The Cost of Raw Materials is usually quoted in pounds, 
therefore it is necessary to find volume or weight of stock. 
Cast iron weighs 0.26 lbs. per cu. in. 
Carbon steel weighs 0.28 lbs. per cu. in. 
Bronze weighs 0.3 1 lbs. per cu. in. 
Brass weighs 0.30 lbs. per cu. in. 
In all operations on machinery in a factory, approximately 
10% of time is lost in changing tools, etc. every day. 

Depreciation is the natural decrease in the value of equip- 
ment. 

Maintenance is the amount of money required to maintain 
the equipment in good working condition. 



COST CALCULATION 



87 



EXERCISES 

1. What would be the cost of a bar of high carbon steel 3" in diameter 
10 ft. long? 

2. What would be the cost of a round bar of high speed steel 4" in 
diameter by 10 ft. long? 

3. What would it cost for material to make 24 cast iron spur gears 
1" wide, outside diameter 4", allowing f" for cutting off tools and -|" 
for finishing over all. (Cast iron weighing 0.26 lbs. per cu. in.) 

4. Find the cost of a 1" round piece of brass 12 ft. long estimating 
brass at 0.30 of a lb. per cu. in. 

5. What would a piece of high carbon steel 4" X 4" X 10 ft. long cost? 

6. What would it cost for material for 12 die blocks of high carbon 
steel 1 1" thick, 6" wide by 8" long? 

7. What will be the cost of a taper piece of C.R.S. 10" long, 3" in 
diameter at the large end and 1" in diameter at the small end? 

8. Find the cost of a 1" hex. brass bar 16 ft. long. 

9. Find the cost of 6 pieces of C.R.S. \" X 4" X 12". 

10. What will be the cost of 100 pieces of high speed steel |" X f" 




Bushing 



11. Determine cost of material and machine operations on 60 tool 
steel bushings, finished size 1" diam- 
eter X 2" long X \" hole, rough size 
stock !■£$" diameter allowing \" for 
cut off. Turn with one cut at 0.004" 
feed per rev., and 150 r.p.m. Drill 
at 0.005" feed and ream at 0.015" 
per rev., allowing 30 sec. for facing 
and cutting off. Lost time on each piece 5 sec. for change of turret, 
etc., wages 37! cents per hour. 

12. What will it cost to machine on a turret lathe, 100 spur gear 
blanks made of cast iron 4" outside diameter, 1 \" hole, if" wide, running 

at 70 r.p.m., using a feed of 0.010" per rev. 
for turning outside diameter, 0.005" feed for 
drilling and boring, and 0.020" feed for ream- 
ing, allowing 2 sec. for revolving turret be- 
tween operations, wages 37! cents per hour. 
13. Estimate cost of cutting 30 teeth on 
each gear in the preceeding problem, on a 
B. & S. gear cutter, if it requires 14 sec. to 

make each cut and 1 sec. for return and indexing, allowing 1 minute 

to change blanks, wages 35 cents per hour. 




Gear Blank 



88 



INDUSTRIAL MATHEMATICS 



# 



14. Determine cost of machining and material, doing work on a 
lathe for a bronze bushing 3" outside 
diameter, 3" long and 2" hole finish size, 
rough size 3s" outside diameter and 3I" 
long, i\" cored hole, spindle speed 150 
r.p.m., chucking 30 sec, feed for drilling 
0.005" per rev., for rough and finish bor- 

ing 0.005" feed, reaming 0.020" feed, 
placing on arbor 60 sec, rough and finish turning on outside at 0.006" 
feed, for facing off each e'nd 20 sec, allowing 50 sec for changing each 
tool, wages 37! cents per hour. 

15. Estimate cost of material and machining on Acme automatic 
screw machine for making 10,000, 

\" hexagon head screws 2" long, 
body IjV long. It requires 30 
sec to do all operations on one 
screw. The four operations being 
done at the same time, allowing 
\" for cutting off; wages 45 cents 
per hr., one man running 3 ma- 
chines and |" C.R.S. hex. stock is used. 

16. What will it cost to make 10 pair of tool steels parallels \" X \" 
X 6", finish size, rough size -^" Xn X 6|". Take one cut with a 

0.005" feed per stroke overall for fin- 

— gl_ ^| *\fafc- ishing on shaper. Leaving 0.016" 

overall for grinding and using a cross 

feed of 0.010" and 0.002" deep per 

cut on a surface grinder. It requires 

6 sec. for a cut and return for grinder, 

and 3 seconds for shaper, allowing 3 min. for shaping off each end, and 

4 min. for grinding each end. Lost time — 1/10, wages 35 cents per 

hour. 




2S 



X 12-Hex. Head Screw 






D 



Parallel. 20 Wanted 



Levers 



A Lever is an inflexible rod capable of motion, about a 
fixed point, called a Fulcrum. The rod may be straight, 
curved or bent at any angle. 

There are 3 kinds of levers, or in other words 3 ar- 
rangements of the force, weight and fulcrum. 



LEVERS 



In the Lever of the First Class, the fulcrum lies between 
the points at which the force and load act. Fig. I. 



*-W.At- 



- P. A.— 
Fig. I 



4? 



WA.--X- PA: 



— »<p 



Fig. II 



W A. 



Ar-*|f 






k--PAr-HP 



Fig. Ill 



In the Lever of the Second Class, the load acts at a point 
between the fulcrum and the force. Fig. II. 

In the Lever of the Third Class, the action of force is 
between the load and the fulcrum. Fig. III. 

Levers are usually used to 
gain power at the expense of 
time, thus, in a first class 
lever, if the distance from the 
fulcrum to power is 5 times the 
distance to the weight, it will give 5 times the power, but it 
will take a movement 5 times greater than the weight moves. 

Levers of the 3d class involve a mechanical disadvantage, 
as the power must always be greater than the weight. 

Law of Levers. The power multiplied by its distance 
from the fulcrum is equal to the weight multiplied by its 
distance from the fulcrum. 

The law for bent levers (Fig. IV and Fig. V) is the same 
as for straight levers, but the length of arms is computed on 
lines from the fulcrum at right angles to the direction in 
which the power and weight act. 




■WA— >^ PA. 

Fig. IV 



-W.A.— 



Vv. 



P. A. 



Fig. V 



90 INDUSTRIAL MATHEMATICS 

Formulas: 

P = Power or force. 
W = Weight or resistance. 
PA = Power arm or distance from fulcrum to point where 

power is applied. 
WA = Weight arm or distance from fulcrum to point where 

weight or resistance is applied. 
Thus, the law of the levers becomes P X PA = W X WA. 

W XWA P X PA 

P= , W= , 

PA WA 

WXWA P X PA 

PA = , WA = . 

P W 

Example: What force 15" from fulcrum will balance a 
weight of 300 lbs. 5" from fulcrum? 

100 

WXW A 3QQX5 „ ,. , 

P = — , P = = 100 lbs. (Ans.). 

PA 15 

3 

EXERCISES 

1. What force 8 ft. from the fulcrum will balance a weight of 500 
lbs. 10" from the fulcrum in a lever of the 1st class? Of 2d class? 

2. When a weight of 426 lbs. is balanced on the end of a lever of 
the 1st class, by a force of 60 lbs. 2\ ft. from fulcrum, what is the distance 
from weight to fulcrum? 

3. What force is required to raise a 1 ton weight, using a lever of 
the 2d class, if the W.A. = 24" and P.A. = 10 ft. 

4. Find the weight that a force of 270 lbs. will lift with a lever of 
the 3d class if P.A. = 4 ft. and W.A. = 10 ft. 

5. What must be the length of W.A. to lift a 150 lb. weight with a 
lever of the 1st class if P.A. = 6 ft. and power = 50 lbs.? 

6. What must be the length of W.A. if a force of 750 lbs. 22" from 
fulcrum lifts a weight of 1000 lbs. in a 2d class lever? 

7. In a lever of the first class what weight will a 100 lb. force acting 
4 ft. from fulcrum at an angle of 30 deg. lift, if the weight is 20" from 
the fulcrum? 



LEVERS 



91 



8. What will be the force in lbs. applied at the brake shoe, if a 
pressure of 50 lbs. is applied with a foot at the end of arm "A"? 




mwmz 



9. What will be the force in lbs. exerted by rod "A," if a person 
pulls with a 40 lb. force on the lever as shown in sketch? 





. y/Jy///////, 



10. If an air brake lay-out is as per sketch, what will be the force in 
lbs. exerted against the brake shoe, if the air cylinder is 10" in diameter 
and the air pressure is 100 lbs. per sq. inch? 



Pulleys 

A Pulley is a wheel mounted to revolve on an axis and 
having a grooved rim in which a cord, band or chain is 
passed to transmit the force applied in another direction. 



92 



INDUSTRIAL MATHEMATICS 



A Pulley Block is a device for holding one or more puiicy 
as a unit. 





Fig. I 



Fig. II 



Fig. Ill 




Pulleys are either Fixed or Moveable, depending upon 
whether they are held in a fixed position or move with the 
given load. 

Fixed pulleys are those that have a fixed block Fig. (I) 
and are generally used to change the direction of the power 
applied. 

Moveable pulleys are those that have movable centers 
(Fig. II). 

Fig. Ill shows a combination of a fixed and movable 
pulley. In this arrangement, in order for the weight (W) 
to move i ft. the power (P) must move through a distance 
of 2 ft., thus W = 2?. 

Whenever possible the pulleys should be so arranged that 
the pull would come in the direction the weight is to be 
moved, as shown in Fig. IV. 

Rule for Pulleys. — The force (P) multiplied by the number 
of strands (N) from the movable pulley, will equal the 
weight (W) that can be raised, or P X N = W. 

Thus, if a force of I lb. is exerted in Fig. Ill, a 2 lb. weight 
would be raised P X N = W. 1X2=2 (Ans.). 

If a 1 lb. force is exerted at P in Fig. IV, a 3 lb. weight 
would be raised. P X N = W. 1X3 = 3 (Ans.). 



PULLEYS 



93 



Fig. V shows a Differential Pulley arrangement used 
extensively in the machine shop. 
In this form of pulley an endless 
chain replaces the rope. The two 
pulleys at the top are of slightly 
different diameters, but rotate to- 
gether as one piece. In operation, 
as the chain is drawn over the 
large wheel, it passes around the 
lower pulley, up over the small 
wheel from which it is unwound, 
causing the loop in which the 
movable pulley rests to be short- 
ened by an amountequal to the 
difference in circumference of the 
two upper wheels, when they have made one revolution. 
This would cause the weight to raise one-half of this amount. 

Example: In Fig. V the two upper pulleys are respectively 
16" and 15" in circumference. As the power applied moves 
through a distance of 16" the small pulley will unwind 15" 
of chain, causing a shortening of the loop c of 1", which 
will raise the weight (W) \" , giving a ratio of load to power 
of 32 to 1. 




Fig. V 



Rule. 



W(R - r) 
2R 



W = 



2PR_ 

R - r 



EXERCISES 

1. How many movable pulleys would be required to balance 100 lbs. 
with a 50 lb. weight? 

2. Determine weight required to balance a 1 ton weight, using 5 
movable pulleys arranged as in Fig. IV? 

3. What power is required to raise a weight of 150 lbs., using pulley 
arrangement in Fig. II? 

4. Using pulley arrangement in Fig. III? 
5- Using pulley arrangement in Fig. IV? 



94 INDUSTRIAL MATHEMATICS 

6. What weight can be raised with a force of 80 lbs., if 3 loose pulleys 
are used, arranged as in Fig. III? 

7. What power will be required to raise a weight of two tons with a 
differential pulley, if the diameter of the two upper pulleys are 19" and 
20" respectively? 

8. What weight can be raised with a force of 100 lbs. with a differ- 
ential pulley, if the two upper wheels have 18 and 17 teeth respectively, 
2 pitch, (distance from one tooth to another = 1.5708"). allowing 20% 
for friction? 

9. Using a differential block with pulley 15" and 12" in diameter, 
what force is required to raise a 250 lb. casting where there is a friction 
loss of 15%? 

10. In a differential block with pulley 10" and 15" in diameter, a 
pull of 90 lbs. is required to raise a weight of 500 lbs. How much 
force is used up in overcoming friction? 

Screws 

A Screw is a modified form of inclined plane. The lead 
of the screw or the distance the thread advances in going 
around the screw once being the height of the incline, and 
the distance around the screw measured on the thread being 
the length of the incline. 

When a force is applied to raise a weight or overcome 
resistance by means of a screw or nut, either the screw or the 
nut may be fixed, the other being movable. 

The force is generally applied at the end of a wrench or 
lever arm, or at the circumference of a wheel. 

The ratio of the power to weight is independent of the 
diameter of the screw. 

In actual work a considerable proportion of the power 
transmitted is lost through friction. 

Rule. — The force applied multiplied by the circumference 
of the circle through which the force arm moves, equals the 
weight or resulting force multiplied by the lead of the screw 
in inches. 



SCREWS 



95 



Formulas 

P : W : : L : 2wR. 

P = Power applied. 

L = Lead of screw. 

R = Length of bar, wrench or 
radius of hand wheel used 
to operate the screw. 

W = Resulting force or weight 
moved. 

W X L 
P = , 

2irR 

EXERCISES 




w = 



P X 2ttR 



i. What weight can be raised with a 4 pitch jack screw, if a force 
of 75 lb. is applied to a lever 15" long? 

2. What pressure can be obtained with a |" X 12 screw, if a 25 lb. 
force is applied at the end of a wrench 8" long? 

3. What length of wrench is required to obtain a pressure of 113 10 lbs. 
on ai" X 8 screw, if a 25 lb. pressure is exerted on the end of the wrench? 

4. If a pressure of 125 lbs. is applied to a 14" lever on a 4 pitch jack- 
s crew, estimate pressure if 20 percent is lost through friction. 

5. Estimate pressure produced on a milling machine vise if the screw 
has 6 threads per inch, length of handle 10" and pressure applied is 75 lbs. 
loss through friction 30 percent. 

6. Estimate pitch of thread required to give a pressure of 3770 lbs., 
if a 25 lb. pressure is applied to a lever 12" long. 

7. Determine leverage required for a screw of 8 pitch, to give a pres- 
sure of 5000 lbs. at end of screw, if a pressure of 25 lbs. is applied at 
ond of lever, and loss through friction is 30 percent. 

8. How many jack screws must be used to raise a machine weighing 
14844 lbs., if the screws have a \" lead, io§" lever and 25 lb. pressure 
applied to lever, allowing 25 percent loss through friction? 



Inclined Planes 

The Inclined Plane is a flat surface sloping or inclined 
from the horizontal. A body moving up an inclined plane 



9 6 



INDUSTRIAL MATHEMATICS 



as opposed both by gravity and friction, while one moving 
down an inclined plane is assisted by gravity and is opposed 
by friction only. 

When the force which is being applied, is exerted in a 
direction parallel to the inclined surface, as in Fig. I, it is 
evident that the power must move through the distance 
equal to the incline in order to raise a weight the desired 
height. The gain in power will then be equal to the length 
of the incline divided by the height. 

Rule.— P :W::H:L. 



P = 



W XH 



W = 



P X L 
H 





If the force acts along a line parallel to the base B, as in 
Fig. II, then P : W : : H : B. 



WXH 



W = 



P X B 



B H 

If the force acts at any angle to the plane as X in Fig. Ill, 
then P : W : : sin Y : cos X. 

100 







Fig. Ill 



100' 
Fig. IV 



30 



INCLINED PLANES 97 

W X sin Y P X cos X 

P = •. W = — — . 

cos X sin Y 

Grade Percent or percent grade is the ratio of the height 
or elevation to the space covered, i.e., if a road elevation 
rises I ft. in ioo ft. it is called a i% grade. If a road rises 
30 ft. in 100 ft. it is a 30% grade. If it rises 100 ft. in 100 ft. 
it is called a 100% grade, or equivalent to an angle of 45 
degrees (Fig. IV). 

EXERCISES 

(In the following problems, friction will not be considered) 

1. In Fig. I, what force is required to roll a 1 ton weight up the 
incline if B is 18 ft., and H 8 ft.? 

2. In Fig. II, what force is required to roll a 1 ton weight up the 
incline if B is 18 ft., and H 8 ft.? 

3. What weight can be drawn up the incline in Fig. I, if power is 
1800 lbs., H is 4 ft. and B 20 ft. 

4. What weight can be drawn up the incline in Fig. II, if power is 
150 lbs., H is 18", and L is 72"? 

5. What power will be required to hold a ball weighing 380 lbs. in 
Fig. Ill, if the incline is 27 deg. 26' from the horizontal, and the force 
is acting 18 deg. 45' from the incline? 

6. What weight will a force of 200 lbs. acting at an angle of 30 deg. 
from the horizontal, sustain if the weight is on an incline whose base 
is 18 ft. and height 4 ft.? 

7. What force will be required to pull a 2200 lb. automobile up a 
15 percent grade, if the force is applied parallel with the grade? 

8. If an automobile going a distance of one mile rises to an elevation 
of 264 ft. What is the average percent grade of the road? 

9. A car with a 122" wheel base is standing on a road with its front 
wheels 18.3 inches higher than the rear wheels. What is the grade 
percent. 

10. What is the grade percent of a roller coaster track, if the angle 
of rise is 24 degrees? 

Wedges 

A Wedge is a pair of inclined planes united at their bases 
(Fig. I). 



98 INDUSTRIAL MATHEMATICS 

The power is usually applied by a blow of a heavy body 
or by pressure. 

Wedges are used for splitting logs and stones, and raising 
heavy weights short distances. Due to the excessive friction 
of wedges, they are not very efficient. 
Rule.— P :W::T:L. 

P = Power applied, 
W = Weight or resistance, 
T = Thickness of wedge, 
Fig. I L = Length of wedge. 

P X L W X T LX P 

W = , L = , T = . 

T P W 

EXERCISES 

(In the following problems, friction will not be considered) 
i. What force will be required behind a wedge used to raise a weight 
of ioo lbs. 3" high, if the wedge has a sliding motion of 15"? 

2. What length wedge must be used to raise a weight of 1 ton with a 
force of 200 lbs., if the thickness of the wedge is 2"? 

3. What weight can be raised with a force of 300 lbs. acting on a 
wedge 10" long and \" thick? 

4. What force will be required to drive a wedge 4" long and f " thick 
to raise a 50 lb. casting? 

5. What force will be required to drive a wedge 12" long and \" 
thick into a log that has a resistance of 1800 lbs. against splitting? 

Gearing Definitions, etc. 

The Center Distance of a pair of gears is the shortest 
distance between the centers of the shafts, on which they 
are mounted. 

The Pitch Circles of a pair of gears have the same diameters 
as a pair of friction rolls which would fill the same center 
distance and revolve at the same velocity ratio. 

The Pitch Diameter of a gear is the diameter of its pitch 
circle. 




Automatic Siur and Be\el Gear Cutting Machine 




Compound Rest 



GEARING DEFINITIONS, ETC. 99 

The Diametral Pitch is the number of teeth a gear has per 
inch of pitch diameter. To find the D.P., divide the number 
of teeth by the P.D. The P.D. in turn may be found by 
dividing the number of teeth by the D.P. 

The Circular Pitch is the distance from the center of one 
tooth to the center of the next, measured along the pitch 
line. To find the circular pitch, divide the pitch circle by 
the number of teeth, or divide ir by the diametral pitch. 

The Size of Gear Tooth is designated by its pitch, thus, 
a 10 pitch tooth has an addendum of i/io" and a dedendum 
of i/io". 

The Tooth Thickness is measured along the pitch line and 
is one-half the circular pitch. 

The Addendum is the height of the tooth above the pitch 
line. 

The Dedendum is the depth below the pitch line to which 
the tooth of the mating gear extends. 

The Working Depth is the depth in the tooth space to 
which the tooth of the mating gear extends, and is equal to 
the dedendum plus addendum. 

The Clearance is the distance from the point of the tooth 
to the bottom of the space in the mating gear. 

The Whole Depth is the distance from the top of the tooth 
to the bottom of the same tooth and consists of the addendum, 
dedendum and clearance. 

The Outside Diameter is found by adding twice the adden- 
dum to the pitch diameter. 

The Root Diameter is the diameter at the bottom of the 
tooth space. 

The Face of the gear tooth is that part of the tooth outline 
which extends above the pitch line. 

The Flank is that part of a gear tooth outline below the 
pitch line. 

The Fillet is the rounded corner where the flank of the 
tooth runs to the bottom of the tooth space. 



100 INDUSTRIAL MATHEMATICS 

The Base Circle is the circle from which the involute curve 
is generated. It is drawn tangent to the pressure line. Its 
position will vary according to the pressure angle used. 
The two pressure angles used are the I4§ and 20 deg., the 
latter being used where a short stubby tooth is required 
called the "Stub Tooth," while the former is used to the 
greatest extent. For a 14J deg. pressure angle tooth gear, 
the base circle will lie inside of the pitch circle a distance 
equal to 1/60 of the P.D. 

Rotary Motion can be transmitted by belts, chains, shafts, 
universal joints, friction discs, gearing, etc. 

A correctly cut gear will transmit a uniform motion. 

The Principal Styles of Gearing are spur, herringbone, 
bevel, spiral or helical and worm. 

The two standard tooth curves are the Involute and 
Cycloidal. 

An Involute Curve is most desirable because it will allow a 
certain amount of variation in the center distance and is 
used almost universally. 

The involute curve is generated by unwinding a string 
from the base circle and allowing a point on the string to 
describe a curve, which will be an involute. 

The Cycloidal will not permit any variation in the center 
distance and can be generated by two different circles, 
Epicycloidal by revolving a circle on the outside of base and 
Hypocycloidal on the inside. 

The ratio of the speeds of two gears that run together is 
called their Velocity Ratio, and is in inverse proportion to 
their sizes or P.D. 

Spur Gears are used to transmit power between two shafts 
running parallel with each other. 

Herringbone Gears conform to two spiral gears, one right 
hand and the other left hand, fastened to each other, thus 
equalizing the side thrust. These gears are very quiet in 



SPUR GEARING 



101 



action, due to some part of the tooth always being in full 
action. 

Bevel Gears are used to transmit power from one shaft to 
another when the axes are not parallel to each other, but in 
the same plane. 

A bevel gear blank is similar to a frustum of a cone. 

Spiral or Helical Gears are the same as spur gears, but 
teeth are cut other than at right angles with the axis. 



No. of Cutter 


No. of Teeth 


No. of Cutter 


No. of Teeth 


I 


135 to a rack 

55 " 134 
35 " 54 
26 " 34 


5 . 


21 to 25 
17 " 20 
14 " 16 
12 " 13 


2 


6 . 


3 


7. . 


4 


8 









A Worm and Worm Wheel is used where a greatly reduced 
speed ratio is desired and consists of a single or multiple 
thread worm meshing into a worm wheel similar to a concave 
faced spur whose teeth are cut angular. 

The relative efficiency of different styles of gearing are as 
follows: 1st — spur; 2d — herringbone; 3d— bevel; 4th — 
spiral or helical; 5th — worm. 

For cutting smooth running in- 
volute gear teeth, 8 cutters are re- 
quired for each pitch. These cut- 
ters are adapted to cut from a gear 
of 12 teeth to a rack. The follow- 
ing table shows the list of cutters 
and the number of teeth it will cut: 

Spur Gearing 

Spur gears are the most com- 
monly used gears. They are cy- 
lindrical in shape and the teeth 
are cut parallel with the axis. Fig. I 




102 



INDUSTRIAL MATHEMATICS 



The different terms commonly used for parts of a spur 
gear are as per sketch and explanation. 



In Fig. II and III 

A = Cir. pitch or distance from center of one tooth to next, 
measured on the pitch line. 

B = Clearance. 

C — Addendum — top of tooth be- 
tween O.D. and P.D. 

D = Dedendum — bottom of tooth 
between P.D. and clearance. 

E = Whole depth — addendum, de- 
dendum and clearance. 

F = Working depth — addendum and 
dedendum. 

G = Thickness of tooth — width of tooth from outside to out- 
side on pitch line. 

H = Outside diameter. 

I = Pitch diameter or the diameter of gear from one pitch 
line to the opposite on center line. 




Fig. II 




"/nvo/ufe — ^C^^J-"' 
Fig. Ill 



SPUR GEARING I03 

Abbreviations Generally Used 

P = Diametral pitch, or pitch. 
O.D. = Outside diameter. 
N = No. of teeth. 
Np = No. of teeth in pinion. 
Ng = No. of teeth in gear. 
N.R. = No. of teeth in rack. 

L = Length of rack. 
P.D. = Pitch diameter. 
CD. = Center distance. 
C.P. = Circular pitch. 
Wh.D. = Whole depth. 
YVg.D. = Working depth. 
Add. = Addendum. 
Ded. = Dedendum. 
C = Clearance. 
Th. = Thickness of tooth. 
R.D. = Root diameter. 

Formulas 

P = 7T -r- C.P. or N + P.D. 
O.D. = (N +'2) ■*- P or (N + 2) X C.P. -r tt or P.D. 

+ 2 add. 
C.P. = t + P or P.D. X tt -=- N. 
P.D. = N +■ P or N X C.P. ^ tt or O.D. - 2 add. 
CD. = (Ng + iV£) -f- 2P or (Ng + iV» X C.P. 4-6.2832. 
Clear. = 0.157 + P or C.P. 4- 20. 
Add. = 1 t P or C.P. + tt or C.P. X 0.318. 
Ded. = 1 v ? or C.P. vior C.P. X 0.318. 
Wh.D. = 2.157 vPor 0.6866 X C.P. 
Th. = 1.5708 -rPor C.P. -r- 2. 
N = P X P.D. or tt X P.D. -v- C.P. 
L = x X N.R. -v- P or N X C.P. 
R.D. = O.D. - 2 Wh.D. or P.D. - 2 (Ded. + C). 



104 INDUSTRIAL MATHEMATICS 

EXERCISES 

i. Determine the diametral pitch of a gear whose circular pitch is 
0.5236. 

2. Find addendum, dedendum and clearance of a 4 pitch gear. 

3. What is the length of add. if P.D. of gear is 2" and the gear has 
20 teeth? 

4. Estimate clearance on a 6 pitch gear. 

5. Determine center distance of a 25 tooth and a 60 tooth gear, 
of 10 pitch. 

6. What is center distance of two gears of 40 and 60 teeth, 10 P. 

7. Find outside diameter of a gear of 30 teeth, pitch diameter 5". 

8. Find outside diameter of a gear of 40 teeth and circular pitch 
0.3141- 

9. How many teeth are there in a gear of 4 pitch, 8" pitch diameter? 

10. Estimate O.D. of a gear whose circular pitch is 0.500" and 
N = 60. 

11. Find the length of a rack with 30 teeth, of 6 pitch. 

12. Find the length of a rack with 60 teeth, of 10 pitch. 

13. Find addendum, dedendum and clearance of a 7 pitch gear. 

14. What is the P.D. of a gear 4" O.D. and 8 pitch. 

15. Find thickness of tooth of a 6 pitch gear. 

16. What is the O.D. of a 40 tooth gear, 4" pitch diameter? 

17. Find the clearance of a 10 pitch gear. 

18. Find the pitch diameter of a 30 tooth gear, of 6 pitch. 

19. Give all measurements accurate and complete for a toolmaker 
who wishes to make a master gear of 10 pitch, pitch diameter 5". 

20. Given center distance 5", ratio 2 
to 3, pitch 10; find pitch diameter, out- 
side diameter and number of teeth in each 
gear. 

21. Given approximate center distance 
5|", ratio 15 to 26, 8 pitch; find pitch 
diameter, outside diameter and number 
of teeth in each gear. 

22. If two parallel shafts (as per 
sketch) with a center distance of 3-535" 
are connected with a jack shaft at right 
angles and parallel to the others, the jack 
shaft having two gears on it with a ratio 

of 2 to 1 and the driving shaft having a 20 T, 10 P gear, what will be 
the P.D. and O.D. of the gears on the driving shaft, jack shaft and 




BEVEL GEARING 105 

driven shaft, providing the ratio between the driving shaft and driven 
shaft is 3^ to 1 and the jack shaft is equal distance from the driving 
shaft and driven shaft? 

23. What number of cutter should be used for cutting a 24 tooth 
gear; a 40 tooth gear? 

Bevel Gearing 

Bevel Gears are used to transmit positive rotary motion 
to shafts at an angle to each other, and in the same plane. 

The teeth of a bevel gear are made on a frustum of a cone 
whose apex is the same point as the intersection of the axes 
of the shafts. 

Bevel gears usually connect shafts running at right angles. 

When the angle of the shafts is 90 deg. and the velocity 
ratio is 1 to 1, then both gears are of the same size, and are 




Fig. I 



called Miter Gears. If the velocity ratio between two gears 
is other than 1 to 1, the smaller gear is called the Pinion. 

When the pitch of two gears is the same, they will mesh 
properly regardless of number of teeth, providing they have 
twelve or more teeth. 

A gear with less than 12 teeth must be cut special to avoid 
interference of teeth while rolling. 

The P, O.D. and P.D. of a bevel gear are always reckoned 
on large end of tooth. 



106 INDUSTRIAL MATHEMATICS 

Formulas 

Tang, of P.C. ang. of pinion = Np -r- Ng. 

Tang, of P.C. ang. of gear = Ng -r- Np. 

Pitch diameter = N -*- P. 

Addendum = i -4- P or C.P. X 0.318 or 

C.P. + ir. 
Dedendum = 1 -f- P or C.P. X 0.318 or 

C.P. + T. 
Whole depth of tooth = 2.157 + P or C.P. X 0.687. 

Pitch cone radius = P.D. -r- (2 X sin P.C. ang.). 

Thickness of tooth = 1.571 v P or C.P. •¥ 2. 

Small addendum = (P.C.R. - B) -J- P.C.R. X 

add. 
Small thickness of tooth = (P.C.R. - B) H- P.C.R. X 

thick. 
Angle of add. = Add. -5- P.C.R. = tang. 

Angle of ded. = Ded. -4- P.C.R. = tang. 

Face angle = 90 deg. — (P.C. ang. + add. 

ang.). 
Cutting angle = P.C. ang. — ded. ang. 

Angular addendum = Cos. of P.C. ang. X add. 

Outside diameter = Ang. add. X 2 + P.D. 

No. of teeth for which to N 

select cutter. Cos. of P.C. Ang. 

P.C. rad. = Pitch cone radius = A 

W. of F. = Width of face = B 

Ang. add. = Angular addendum = C 

Add. ang. = Addendum angle = D 

Ded. ang. = Dedendum angle = E 

P. line = Pitch line = F 

P.C. ang. = Pitch cone angle = G 

Cut. ang. = Cutting angle = i7 

O.D. = Outside diameter = / 



BEVEL GEARING 



I07 



P.D. 

P.C. ang. 
P.C. ang. 
Wh. D. 
Add. 
Ded. 

E. ang. 

F. ang. 
Ng 
Np 

N 
P 
T 
N' 



= Pitch diameter = J 

G. = Pitch cone angle of gear = K 
P. = Pitch cone angle of pinion = L 
= Whole depth = M 

= Addendum = N 

=. Dedendum = 

= Edge angle = P 

= Face angle = Q 

= No. of teeth in gear 
= No. of teeth in pinion 
= No. of teeth 
= Diametral pitch or pitch 
= Thickness of tooth 
= No. of teeth for which to select cutter 



yC/ecrrcrnce 

r / 




-----V--"" 



Fig. II 



EXERCISES 

(In figuring these problems, make the width of face equal to \ of the 
pitch cone radius) 
1. Find P.C. rad. of a bevel gear whose P.D. is 4", P.C. ang. 60 deg. 
10 pitch and 40 teeth. 



108 INDUSTRIAL MATHEMATICS 

2. Estimate cutting angle of the above gear. 

3. Estimate width of face of the above gear. 

4. Find addendum angle of the above gear. 

5. Find dedendum angle of the above gear. 

6. Find outside diameter of the above gear. 

7. Find pitch diameter of the above gear. 

8. Find whole depth of tooth for a 10 pitch gear. 

9. Determine the thickness of cutter at pitch line to use for the 
first cut on a bevel gear of 30 teeth, 6 pitch and 60 deg. P.C. ang. 

10. At what angle from horizontal would you set the dividing head 
to cut a bevel gear, P.C. ang. 48 deg., 6 pitch and 30 teeth. 

n. Estimate face angle of a bevel gear with 30 teeth, 6 pitch and a 
P.C. ang. of 60 deg. 

12. Determine the whole depth of tooth at small end of a bevel gear 
with 30 teeth, 6 pitch and P.C. ang. of 54 deg. 

13. Find the P.D. at the small end of a 40 tooth gear, 8 pitch and a 
P.C. ang. of 42 deg. 

14. Find the outside diameter of a bevel gear with 42 teeth, 6 pitch 
and a P.C. ang. of 38 deg. 

15. Give measurements accurate and complete for a master bevel 
gear of 50 teeth, 10 pitch and 60 deg. P.C. ang. 

16. If two 6 pitch bevel gears, shafts at 90 deg., have a velocity ratio 
of 3 to 5, with 18 teeth in the pinion, what are the O.D. and face angles 
of the blanks? 

17. In a pair of 2 pitch bevel gears, with shafts at 90 deg., having 
a velocity ratio of 2! to 1, the pinion has 24 teeth. Find the face angle, 
pitch cone angle and cutting angle of both gears. 

18. Determine the number of cutter required to cut the bevel gear 
in problem No. 9. 

Worm Gearing 

Worm Gearing is used to transmit power between two 
shafts at 90 to each other, but not in the same plane, and 
is generally used when it is desired to obtain smoothness of 
action and great speed reduction from one shaft to another. 

The greatest objection to worm gear drives is the excessive 
sliding friction between the teeth, thus making them very 
inefficient and subject to heating. 

A Worm is a screw so cut as to mesh properly with the 



WORM GEARING 



109 



teeth of a worm wheel, the included angle of the sides being 
29 deg. 

The Worm Wheel is similar to a spiral spur gear. It 
usually has a concave face and 
the tooth spaces are concave and 
at an angle other than 90 deg. to 
the side of the gear. These teeth 
are generally cut by first being 
indexed and gashed, and after- 
wards cut to true form with a 
hob. But if possible they should 
be cut on a gear hobbing machine, 
for accurate results. 

A Hob is a cutter slightly larger 
in diameter than the worm, and 
it appears somewhat like a worm 
with the exception that flutes are 
cut into it to form the cutting teeth. 

The Linear Pitch is the distance from the center of one 
tooth to the center of the next, measured on the pitch circle. 

The Lead sometimes differs from the pitch and it is the 
distance a tooth on the worm would advance in one revolu- 
tion, or the distance the worm wheel advances in one com- 
plete turn of the worm. 




Fig. I 



Formulas for the Worm 

Lead = linear pitch X no. of separate threads on the worm. 

Linear pitch = lead -5- no. of separate threads on the worm. 

Addendum = linear pitch X 0.3183. 

Whole depth of thread = linear pitch X 0.6866. 

Width of threading tool at end or width of bottom of space 

= linear P. X 0.31. 
O.D. = P.D. + (2 X add.). 
P.D. = O.D. - (2 X add.). 



no 



INDUSTRIAL MATHEMATICS 



P.D. = (2 X center distance) — P.D. of gear. 
Root diameter = O.D. — (2 X whole depth of tooth). 
Co-tangent of angle of worm tooth or gashing angle of wheel 
= (P.D. X t) -r- lead. 

Formulas for Worm Wheel 

P.D. = (no. of teeth in gear X linear pitch of worm) 4- ir. 
Throat diameter = P.D. of worm wheel + 2 X add. 
Radius of throat = \ of O.D. of worm — (2 X add. of worm). 
Center distance = (P.D. of worm + P.D. of gear) 4- 2. 
O.D. = (throat radius — throat radius X cosine of \ face 
angle) X 2 -f- throat diameter of wheel. 



Worm 

A = Clearance. 

B = Working depth of tooth. 

C = Whole depth of tooth. 

D = O.D. of worm. 

E = P.D. of worm. 

F = Ang. of helix. 

G = Linear pitch. 




P. Line 



Fig. Ill 




Worm Thread 

larged view) 

A = Clearance. 

B = Addendum. 

C = Dedendum. 

H — Lead. 

I = Thickness of end of tool or bottom of space. 

J = 2 an g- of tooth. 

K = Root diameter of worm. 



WORM GEARING 



III 




Worm Wheel 



Worm Wheel 

A — O.D. of worm wheel. 

B = Center distance of worm 

and worm wheel. 
C — Ang. of face. 
D = Throat radius. 
E = Pitch diameter. 
F = Throat diameter. 
G = Clearance. 

EXERCISES 

i. Determine pitch diameter of worm 
wheel — number of teeth in wheel 30 and 
linear pitch 0.200". 

2. Find throat diameter of a worm 
wheel of 40 teeth, linear pitch of worm 0.230". 

3. Find outside diameter of a worm wheel whose face angle is 70 
degrees, throat radius is §", number of teeth 32, and linear pitch of 
worm 0.200". 

4. Find center distance of a worm gear of 48 teeth, linear pitch of 
worm 0.200", outside diameter of worm i\". 

5. Find angle of worm tooth or gashing angle of wheel, if the outside 
diam. of the worm is i£", linear pitch 0.240", with a double thread. 

6. Find root diameter of a worm whose outside diameter is i|" 
and linear pitch 0.175". 

7. Determine thickness of thread tool at small end for a worm 
whose linear pitch is 0.173". 

8. Estimate velocity ratio of a worm and worm gear if the gear has 
30 teeth. (Worm has a triple thread.) 

9. Find radius of curvature of the worm wheel throat, if the pitch 
of the worm is 0.150" and the outside diameter is 1". 

10. Determine all measurements accurate and complete for a mechanic 
to make a set of master worm gears; linear pitch of worm 0.200", double 
thread, outside diameter of worm i|", number of teeth in wheel 32, 
and a face angle of 70 deg. 

Spiral Gearing 

Spiral or Helical gears are used to drive shafts at angles 
to each other but not in the same plane. 



112 



INDUSTRIAL MATHEMATICS 



The Velocity Ratio depends upon the number of teeth 
and their helical angles. 

The velocity ratio of two spiral gears is proportional to 
their pitch diameters only when the spiral angles of the gears 
are the same; or 45 deg. when the axes are at 90 deg. to 
each other. 

The sum of the Spiral Angles of two mating gears must 
always equal the angle between the shafts. 

The Normal Pitch is the pitch measured at right angles 
with the tooth. 

The Tooth Angle is the angle the tooth makes with the 
axis of the gear. 

The Center Angle of a pair of spiral gears is the angle 
made by the axes of the gears. 

The Normal Diametral Pitch is the D.P. of the cutter 
used for cutting the teeth in the spiral gear. 

The Thickness of Cutter at the pitch line for milling spiral 
gears should equal \ of the normal circular pitch. 



(Abbreviations Generally Used in Spiral Gearing) 

a = small gear. 
A = large gear. 
a = spiral angle. 

r = center angle (or ang. between 
shafts). 
Pn = normal diameter pitch (pitch 
of cutter). 
N = no. of teeth. 

N' = no. of teeth for which to se- 
lect cutter. 
C = center distance. 
L = lead of tooth helix. 
Tn = normal thickness of tooth at 
pitch line. 




Fig. I 



SPIRAL GEARING 



113 



Add. = addendum. 
Ded. = dedendum. 

W = whole depth of tooth. 
D = pitch diameter. 
= outside diameter. 
V = velocity of large gear. 
v = velocity of small gear. 
T — thickness of cutter. 
K = no. of teeth for which to select cutter for large gear. 

The Number of Cutter to use for cutting spiral gears is 
not selected with reference to the actual number of teeth in 
the spiral gear, but of an imaginary spur gear arranged at 
right angles to the normal pitch. 

Formulas 



r = OLA + OL a . 

N 
D = 



C = 



Pn cos a 
Da + DA 

2 
tD X cot a. 



N 
N' = 

(cos a)' 



Add. = 



Pn 



Ded. 


= 


Pn' 


W 


= 


2.157 

Pn 


Tn 


_ 


I.57I 




Fig. II 



Pn 



114 INDUSTRIAL MATHEMATICS 

= D + 2B. 

v : V : : £>^4 X cos au : Da X cos a a . 

N 
K = 



Normal pitch = 



(cos a) 3 
P 



cos of tooth ang 



EXERCISES 



i. Two 25 tooth spiral gears, with a velocity ratio of 1 to 1 and shafts 
at 90 deg. are to be cut with a 10 P. cutter. Find P.D., O.D., whole 
depth, add, center distance and lead of tooth helix. 

2. Two spiral gears with axes at 90 degrees, velocity ratio 2 to 3, 
are to be cut. The approximate P.D. of small gear is 2j", and its 
tooth angle is 30 deg., and normal pitch is 10. Find the pitch diam., 
no. of teeth, O.D., and no. of cutters for both gears, and also center 
distance. 

3. Find the O.D. of two spiral gears that have 18 teeth, with a velocity 
ratio of 1 to 1, cut with an 8 pitch cutter, shafts at 90 deg. 

4. What will be the center distance between 2 spiral gears with a 
velocity ratio of 1 to 1, shafts at 90 deg. each having 80 teeth, if the 
normal pitch is 12. 

5. Two spiral gears are required to have a velocity ratio of 1 to 1, 
shafts at 90 deg. to each other, a 6 pitch cutter to be used with an O.D. 
to be as close as possible to 5". Find P.D., number of teeth, normal 
cir. pitch, center distance and number of cutters to cut the above gears. 

6. Two spiral gears with axes at 90 deg. to each other, velocity ratio 
2 to 1, normal pitch 12, center distance 2j", number of teeth in small 
gear 12, tooth angle 38 deg. 20 min. Find P.D., O.D., cir. pitch, normal 
pitch, and number of cutter for both gears, also number of teeth and 
tooth angle of large gear. 

7. What is the velocity ratio of 2 spiral gears if the large gear has a 
P.D. of 5" and tooth angle of 60 deg. and the small gear has an approx- 



REVIEW EXERCISES 

1. A pole was 2/7 under water, the water rose 8 ft.; and then there 
was as much under water as had been above water before. How long 
is the pole? 



REVIEW EXERCISES 115 

2. A boat goes 16 1 miles per hour down stream and 10 miles an hour 
up stream. If it is 225 hours longer in coming up stream than going 
down, how far down did it go? 

3. Land worth $1000 an acre is worth how much a front foot of 90 
ft. depth, counting off 1/10 for streets? 

4. A machinist took a cut on a lathe off a piece of round cast iron 
2" deep; the piece was 8" in diameter and 6" long. How much did the 
material weigh that was cut away? 

5. Find the length of a minute hand on a clock whose extreme point 
moves 4" in 3 min. 28 sec. 

6. How many 1" balls can be put in a box which measures inside 
5" X 10" X 10" deep? 

7. A12" ball is in the corner where walls and floor are at right angles. 
What must be the diameter of a ball placed in the back of this, and 
which will touch the 12" ball and also the same floor and walls? 

8. If j" of rain fell, how many barrels of water will be caught by a 
cistern which drains a flat roof 52 ft. by 38 ft.? 

9. What peripheral distance would a lathe tool travel on threading 
a piece of work 2" in diameter and 10" long, if the lead is f"? 

10. Four holes are required in a jig as per sketch below. Find dimen- 
sions (xi, X2 and ^3). 

11. If 15 men cut 480 cords of wood in 10 A <NSs v& 
days of 8 hours each, how many boys will it * / | N. 
take to cut 1152 cords of wood only 2/5 as &/ /i Y~ 
hard, in 16 days of 6 hours each, provided y | x %V | *h* 
that while working, a boy can do only f as / / 
much as a man, and that \ of the boys are I _ 7 « I 7^~f"z*"I] 
idle at a time throughout the work? 8 

12. Extract the cube root of 141 72488. 

13. Extract the cube root of 2.6 (to three decimals). 

14. What would a taper piece of cast iron weigh, which was 10" long, 
f" in diameter at the small end and 3" in diameter at the large end? 

15. How many pounds of round cast iron stock would you order to 
make the above piece without allowing for waste on ends and on large 
diameter. 



PART III 



Dovetail Slides 

In the shop dovetail slides are usually measured by placing 
round plugs or rods of such diameter that they will bear on 
the angular surface and then measuring overall or the 
distance between the plugs or rods. 

Dovetails are generally dimensioned as shown in Fig. I, 
and the above method is used for checking the dimensions 
as in the shop it is impossible to measure accurately the 
overall dimensions from edge to edge due to these edges or 
measuring surfaces not being absolutely sharp. 





Fig. I 
To obtain the dimensions X and Y (Fig. II) which are 





k B 



k— a— J 



Fig. II 



necessary for the practical measuring of dovetail slides, the 
following formulas may be used: 

X = A - [Z>(i + cot i4>)l, 
Y = D(i + cot §0) + B. 

Example: Find the distance Y on a dovetail slide if the 

116 



DOVETAIL SLIDES 117 

blue print gives the dimension A as 3", B as 1.846" and the 
angle 6o°, providing plugs 0.750" in diameter are used. 

Formula: Y = D{i + cot J0) + B = 0.750(1 + 1.732) 
-f- 1.846 = 0.750 X 2.732 + 1.846 = 2.049 + 1.846 

= 3.895" (Ans.). 
Example: Find the distance X in the above problem. 

Formula: X = A - [Z>(i + cot §0)] 

= 3 - [0.750(1 + 1.732)] = 3 ~ [o.750 X 2.732] 

= 3 - 2.049 = 0.951" (Ans.). 

EXERCISES 

1. What will be the overall length in measuring a male dovetail, if 
the following data is given on the blue print: angle 70°. width at bottom 
2.886", providing plugs f" in diameter are used. 

2. What will be the distance between to f" plugs placed in a female 
dovetail which measures 2" wide at the bottom and the included angle 
is 6o°? 

3. What will be the length Y in Fig. II, if the angle is 50 degrees, 
D = 0.750" and the width B is 6"? 

4. Find the distance X in Fig. II? If D is |", <£ = 6o°, the depth 
equals one inch and the width A is 2.640". 

5. What should be the distance Y of a properly machined dovetail 
to fit the dovetail in the above problem, if a clearance of 0.002" is allowed 
for a sliding fit, providing \" plugs are used? 

Screw Threads 

There are several different forms of threads such as the 
United States Standard, Sharp V, Square, Whitworth, Acme, 
Briggs pipe, etc. 

The most commonly used threads are the United States 
standard, sharp V, and square threads. 

Standard Sharp V Thread (Fig. I).— The sides of the 
thread form an angle of 60 degrees with each other, and 
is theoretically sharp at the top and bottom. 

1 

p = pitch = 



no. threads per inch 



Il8 INDUSTRIAL MATHEMATICS 

d = depth = p X cos 30 deg. or 0.866 p or 



0.866 



D = outside diameter (Fig. II), 



no. threads per inch 

i p r l i 





Fig. I 



P.D. = pitch diameter = D — d or D 



Fig. II 

0.866 

N 



( 0.866 \ 

R.D. = root diameter = D — {2d) or D — I X 2 I . 

/ = lead = distance the screw advances if turned around one 
complete revolution. In a single threaded 
screw the pitch and the lead are equal. In a 
double threaded screw the lead is equal to 
twice the pitch, etc. (Figs. Ill and IV). 



-I—. 




Fig. III. Single Square Thread Fig. IV. Double Square Thread 

United States Standard Thread (Fig. V).— The sides of 
this thread also form an angle of 60 degrees with each other, 
but the thread is flattened at the top and bottom and this 
flat is equal to § of the pitch. 



SCREW THREADS 



119 



p = pitch = 



no. threads per inch 





Fig. V 



Fig. VI 



d = depth = f X p X cos 30 deg., or 0.6495 X p or 

0.6495 



D = outside diameter (Fig. VI). 
P.D. = pitch diameter = D — d or D 



no. threads per inch 

_ Q-6495 

N 



(0.6495 \ 
X 2 I . 

P 
f = flat = width of flat at top and bottom = - or 

J 8 

1 



8 X no. threads per inch 

The pitch diameter is generally measured directly with a 
thread micrometer, but a 
method for measuring accu- 
rately the pitch diameter by 
means of an ordinary mi- 
crometer and three wires of 
equal diameter, can be used. 
The three wires are arranged 
as shown in Fig. VII, one 
wire being placed in the angle 
of thread on one side of the Fig. VII 




120 



INDUSTRIAL MATHEMATICS 



screw, and the other two on the opposite side, then meas- 
uring over the whole with a micrometer. The reading will 
be according to the following formulas, if the pitch diameter 
is correct. 

For sharp V thread. For U.S.S. thread. 

D = M - 3W + 1732^. D = M - 3W + i.5'i55£. 

M = D - 1.732* + 3^. M = D - 1.5155* + 2>W. 

The Tap Drill size theoretically is equal to the root diam- 
eter of the given tap, but in practice it is always a little larger, 
to prevent excessive strain on the tap, thus increasing pro- 
duction. (See appendix, table VIII.) 

The Briggs standard pipe thread is made with an angle of 
60 degrees. It is slightly rounded at the top and bottom. 
The taper of the thread on the diameter equals 1/16" per 
inch or 3/4" per foot. For number of threads per inch, tap 
drill sizes, etc. (See appendix, table IX.) 

Square Thread (Fig. VIII). — The sides of the square thread 
are parallel and the depth of the thread is equal to the 

width of space between the 
teeth. This space is jtheoretic- 
ally equal to one-half of the 
pitch. It is necessary in prac- 
tice to make the space in the 
nut a trifle wider than the thread so as to have a running 
fit between the screw and nut. 

The Screw Thread Angle (angle of helix) ((f)) Fig. II varies 
with the P.D. and the lead of the screw. 
The following formula may be used: 

lead 





< — 

<~f- 


"P > 




* 

d 


--> 


<-!-> 





Fig. VIII 



P.D. X 7T 
The following graphical method 
may be used: 



= tang, of helix angle. 



-Lead 



Circumference 



LATHE CHANGE GEARS 121 

EXERCISES 

1. Find the depth of V and U.S.S. threads of |" pitch. 

2. Find the depth of V and U.S.S. threads of 1/12" pitch. 

3. What is the R.D. of a f" X 10 U.S.S. thread? 

4. What is the P.D. of a 1" X 8 V thread? 

5. What is the theoretically correct tap drill size for a f" X 11 V 
thread ? 

6. What is the theoretically correct tap drill size for a 5" X9 
U.S.S. thread? 

7. What is the theoretically correct tap drill size for a f" X 16 V 
thread? 

8. What must be the width of flat on a U.S.S. thread having 7 
threads per inch? Also one having 2\ threads per inch? 

9. Find the width of tool and boring size of a 1", \" pitch, square 
thread. 

10. What multiple thread must be cut on a screw so that it will ad- 
vance \" in two-thirds of a revolution, providing the screw has a \" 
pitch, square thread. Also figure the distance the tool must be fed into 
the work. 

11. Give all dimensions necessary for making a master plug gage 
if" X 6 standard sharp V thread. 

12. What will be the micrometer reading of a \" X 12 U.S.S. thread 
if the three wire system is used, providing the wires used are 0.070" 
in diameter? 

13. What will be the correct micrometer reading of a 2!" X 10 V 
thread if the three wire system is used, and the wires are 5/64" in diam- 
eter? 

14. What will be the correct micrometer reading of a 2 ye" X 16 
U.S.S. thread if the three wire system is used, wire being 0.075" in 
diameter. Also find the P.D. and R.D. 

15. Find the helix angle of a \" P. screw, 1" P.D. 

Lathe Change Gears 

In cutting threads on a lathe, the lead screw must be 
taken as the first factor and the main spindle as the second. 
If the lead screw has 5 threads per inch, and the lead screw 
makes five complete revolutions, the carriage will travel one 
inch, or the threading tool will have traveled the same 
distance along the piece to be threaded. If the spindle and 



122 INDUSTRIAL MATHEMATICS 

lead screw are geared one to one, the spindle will make the 
same number of revolutions as the lead screw. Therefore 
the same lead will be cut on the work as on the lead screw. 

If the gear on the spindle is J the size as that on the lead 
screw, the spindle will make twice the number of revolutions 
as the feed screw, the spindle revolving 10 times while the 
tool moves one inch. Therefore 10 threads will be cut. 

If the gear on the lead screw and main spindle are con- 
nected with an idler gear (which does not change the ratio) 
this is called Simple Gearing (Fig. I). 

Sometimes it is not possible to obtain the correct ratio 
with two gears, then two more gears must be put into the 
gear train; this is called Compound Gearing (Fig. II). 

To find the gear ratio between spindle and lead screw in 
simple gearing, the following formula is used: 

Threads per in. of lead screw teeth in gear on spindle stud 
Threads per in. to be cut teeth in gear on lead screw 

To cut 12 threads per inch with a lathe that has a 6 pitch 
lead screw according to the above formula. 

6 i 30 = teeth in gear on spindle. 

— = - or — 

12 2 60 = teeth in gear on lead screw. 

For compound gearing the same formula as above is used, 
except that we divide both numerator and denominator into 
two factors. Thus: 

6 2X3 30 X 40 drivers. 
12 4X3 60 X 40 driven. 

The following are the standard gears put out with a Reed 
lathe having a 5 P lead screw: 25-30-35-40-40-45-50-55- 
60-65-69-70-75-80-90. 

The following are the standard gears put out with the 
Pratt & Whitney lathe, having a 6 P lead screw: 30-40-50- 
60-65-70-75-80-90-95-1 00-1 05-1 10-1 15-120. 



LATHE CHANGE GEARS 



123 



Work. 




Fig. I. Simple Train of Gears for Thread Cutting 



Work.. 



Stud Gear^ 



•{(rTTt Intermediate ~fc = 




Gear 




Fig. II. Compound Gears for Thread Cutting 



124 INDUSTRIAL MATHEMATICS 

EXERCISES 

i. What change gears can be used to cut a 13 P thread with a lathe 
that has a 4 P, lead screw, using a stud gear of 20 teeth? 

2. What change gears can be used to cut a n| P thread when lead 
screw is 5 P, using a 30 tooth gear on stud. 

3. Find the stud gear to be used to cut 18 threads per inch, when 
lead screw is 6 P, and screw gear has 90 teeth. 

4. What pitch thread can be cut with a 6 P lead screw if the drivers 
have 75 and 80 teeth and the driven gears have 50 and no teeth? 

5. What screw gear will be used to cut a 24 P double-thread screw 
when lead screw is 6 P and the stud gear has 60 teeth? 

6. What change gears can be used to cut a 5 P thread when the 
lead screw on the lathe is 6 P? 

7. What change gears can be used to cut a 9 P thread when lead 
screw is 6 P, and the smaller gear in the set has 30 teeth? 

8. The spindle gear in a compound gear train has 25 teeth. On the 
idler stud are two gears, driven 60 teeth and driver 30 teeth, and the 
lead screw has 80 teeth. How many turns does the spindle make for 
one turn of the lead screw? 

9. Using an 80 tooth gear on the lead screw, and a 25 tooth gear on 
the stud, how many threads can be cut per inch if the lead screw has 
8 threads per inch. 

10. Using a no gear on the lead screw and a 75 on the stud, with 
compound driven and driver gears of 50 and 80 teeth respectively, how 
many threads per inch will be cut if the lead screw is 6 P? 

Indexing 

The Dividing Head is used to divide the periphery of a 
piece of work into any desired number of spaces or divisions. 

There are generally three interchangeable plates with each 
dividing head. The following list gives the usual number of 
holes per circle on the three plates. 

Plate Number of Holes in the Various Circles 

1 15-16-17-18-19-20 

2 21-23-27-29-31-33 

3 37-39-41-43-47-49 

Some dividing heads have only one plate, in this case the 




Simple Indexing 




Differential Indexing 



INDEXING 125 

plate has holes on each side as follows: One side — 24-25-28- 
30-34-37-38-39-41-42-43; and on the other side — 46-47- 
49-51-53-54-57-58-59-62-66. 

On all standard dividing heads it requires 40 turns of the 
index crank to revolve the dividing head spindle once. 

Simple Indexing 

To find the Number of Turns of the index crank for any 
number of divisions necessary on the work, divide the num- 
ber of turns required for one revolution of the dividing head 
spindle (40) by the number of divisions wanted. 

N = Number of divisions required. 

R = Number of turns of the crank for a given division. 

40 
R =— . 

N 

Example: Find the indexing required for 50 divisions. 

Solution: R = 40/iV or 40/50 or 4/5 of a revolution. Any 
plate divisible by 5 may be used. In this case, taking the 
20 hole circle. Then 4/5 of 20, or 16 will equal the number 
of holes to be moved in the 20 hole circle, for each division. 

Compound Indexing 

Compound Indexing is sometimes used to obtain divisions 
which cannot be secured by simple indexing. In this method 
the crank is first turned a certain amount in the regular way, 
and then the index plate is also turned either in the same or 
opposite direction in order to locate the index crank in the 
proper position. The back locating pin not being adjust- 
able, is in line only with the outside circle of holes in the index 
plates. Thus, the circles with the 20, 33 and 49 hole circles 
must be used for compounding. 

This method was used extensively in the past, but the 
chances of errors are too great in making the complicated 



126 INDUSTRIAL MATHEMATICS 

indexing moves, and even if properly operated, the spacing 
is more or less inaccurate. Therefore this method is very 
seldom used at present, and has been largely superseded by 
differential indexing. 

To find what circles of holes can be used in compound 
indexing, the following method must be used: 

(a) Set down the number of divisions required and resolve 
into factors. 

(6) Choose at random two circles of holes, subtract one 
from the other and factor the difference. 

(c) Place the factors (a) and (b) above a horizontal line. 

(d) Factor the number of turns of index crank required 
for one revolution of the spindle (40). 

(e) Factor the number of holes in each of the chosen 
circles (b). 

(/) Place the factors obtained in (d) and (e) below the 
horizontal line. 

If all the factors above the line can be cancelled by those 
below the line, the two circles chosen will give the required 
number of divisions chosen. If not, other circles must be 
chosen and another trial made. 

Example: If 50 divisions are required and that circles 20 
and 16 are to be chosen for trial divisors. 

a = 50 = 5X5X2. 
b = 20 — 16 = (4) = 2 X 2. 
^=5X5X2X2X2. 
d = 40 = 2X2X2X5. 

e = (20) =5X2X2. (16) =2X2X2X2. 
7=2X2X2X5X5X2X2X2X2X2X2. 
5X5X2X2X2 1 



2X2X2X5X5X2X2X2X2X2X2 64 



(Ans.1 



Thus, the product of the remaining factors below the line 
= 64. This means that we can index 1/50 of a revolution 



INDEXING 127 

by turning the crank forward 64 holes in the 20 hole circle 
and the index plate backward 64 holes in the 16 hole circle. 
This movement may also be reversed, 64 holes in the 16 
hole circle and the index plate backward 64 holes in the 20 hole 
circle, without affecting the result. 

Differential Indexing 

Differential Indexing is on the same principle as compound 
indexing except that the index plate is revolved by suitable 
gears which connect it to the spiral head spindle. 

The rotary or differential motion of the index plate takes 
place when the crank is turned, which turns the plate either 
forward or backward as -may be required. The result is 
that the actual movement of the crank, in indexing, is either 
more or less than the movement in relation to the index 
plate. 

The differential method cannot be used in connection with 
spiral milling, because the spiral head spindle is geared to the 
lead screw of the milling machine. 

The amount of rotation of the index plate may be regulated 
by the difference in velocity ratios of the change gears. 

Example: Find the indexing required for 81 divisions. 

Solution: By simple indexing the index crank would be 
rotated through 40/81 of a turn for each division, but as there 
is no plate with 81 divisions, the spacing is impossible: 
therefore another fraction is selected whose value is near 
40/81, say 40/84 or 10/21, then a 21 hole circle can be used, 
indexing in this way for 81 divisions, giving 81 X 10/21 
= 810/21 or 38 12/21 complete turns of the index crank 
or 1 9/21 turns less than the 40 required for one complete 
turn of the work. By using gears in the ratio of 1 9/21 
to 1, the index plate will make 1 9/21 revolutions, which 
with 38 12/21 turns of the crank will make the 41 turns 
required. Thus the gears will be in the ratio of 



128 



INDUSTRIAL MATHEMATICS 



9 30 

— or — 
21 21 



6X5 48 X 40 driven 

or - (Ans.) 



7X3 56 X 24 drivers 

If the motion of the index plate must be in the direction 
opposite to the movement of the index crank, idler gears 
must be used. 

The following gears are generally available for differential 
indexing: 24-24-28-32-40-44-48-56-64-72-86-100. 

Angular Indexing 

With the standard index head, where 40 turns of the index 
crank are required, for one revolution of the work, one turn 
of the crank equals 1/40 of 360 deg. or 9 deg. 

Thus, if one complete turn of the index crank equals 9 deg., 
2 holes in the 18 hole circle or 3 holes in the 27 hole circle 
must equal 1 deg., or 1 hole in 18 hole circle will equal \ deg., 
and 1 hole in the 27 hole circle will equal \ of a deg. 





EXERCISES 


I 


What is the simple indexing for 12 divisions? 


2 


« .« .< .. .« .. 2g 


3 


■ ' " " 340 


4 


" " " " " " 85 


5 


" ii5 


6 


" " " compound indexing for 69 divisions? 


7 


tl M <« a i 


" 231 


8 


" " " *' ' 


" 87 


9 


" " " " ' 


" 99 


10 


" " " " ' 


" 147 


11 


" " differential 


" 51 


Z.Z 


<< << << •« < 


" 57 


13 


<< << ii <« « 


" 71 


14 


" " " " ' 


" 101 


15 


" " " ' 


" 352 


16 


" " " angular indexing for 13 degrees? 


17 


.... «: ,« .. .. « „i " 


18 


" ..":'■ " " " 5 deg. 20 min.? 


19 


« 2J «< 3 Q «« 


20 


<< << It u << 


" 7 " 40 " 




Spiral Milling 
A — Gear on worm (driven). 
B — First gear on stud (driver). 
C — Second gear on stud (driven) . 
D — Gear on screw (driver). 



SPIRAL MILLING 1 29 

Spiral Milling 

Spiral Milling is attained by the use of an index head so 
geared to the longitudinal feed screw of the milling machine, 
to impart a rotary motion to the work as it is fed along under 
the cutter by the action of a train of gears. 

The Lead of a Helix or Spiral is the distance, measured 
along the axis of the work, which the spiral makes in one 
full turn around the work. 

By the Lead of the Milling Machine is meant the distance 
the table will travel while the index head spindle makes one 
complete revolution when the gear ratio between the feed 
screw and the worm gear stud is 1 to 1. 

Rule. — Lead of milling machine equals the revolutions of 
the feed screw required for one revolution of the index head 
spindle with equal gears times the lead of the feed screw. 

Lead of spiral product of driven gears 

Lead of machine product of driving gears 

In finding the change gears to be used in a compound train, 
place the lead to be cut, in the numerator, and the lead 
of milling machine, in the denominator, then resolve the 
fraction into its factor and multiply each pair of factors by 
the same number until suitable number of teeth in change 
gears are obtained. 

The following change gears are available on most milling 
machines: 24-24-28-32-40-44-48-56-64-72-86-100. 

Example: Find required gears to cut a 24" lead with a 
10" lead milling machine. 

24 _ 6X4 _ (6 X 12) X (4 X 8) _ 72 X 32 = driven gears 
10 2X5 (2 X 12) X (5 X 8) 24 X 40 = driving gears 

(Ans.). 

The Helix Angle, or angle to which the table is set in spiral 
milling, is found by the following formula: 



130 INDUSTRIAL MATHEMATICS 

x X diameter of work 

tangent of helix angle = . 

lead of work 

A graphical method of determining the helix angle is 

shown in Fig. I. 

f' Draw a base line equivalent 

; to the lead and a vertical line 

I equal to the circumference, then 

c Lead or SA - "*» by connecting these two lines 

Fig. I . , , 

by a hypotenuse and measuring 

the angle (<f>) with a protractor the approximate helix angle 

may be obtained. 

EXERCISES 

i. What change gears are required for a spiral index head to cut a 

48" spiral or lead? 

2. What change gears are necessary to cut a 40" lead? 

3. What change gears are necessary to cut a i§" lead? 

4. What lead will the following gears cut: gear on worm 56. 1st gear 
on stud 28, 2d gear on stud 24, gear on screw 48? 

,5. What lead or spiral can be cut with the following gears: gear on 
worm 40, 1st gear on stud 24, 2d gear on stud 24, gear on screw 32? 

6. What gears and what angle must the milling machine table be 
set at to cut a spiral groove, one complete turn on a piece 8" long and 
2" in diameter? 

7. What angle must a milling machine table be set at, if the following 
gears are used: gear on worm 86, 1st gear on stud 48, 2d gear on stud 56, 
gear on screw 44, to cut a spiral groove, one complete turn on a piece 
2" in diameter? 

8. What gears are required to cut a 60" lead? 

9. What gears are required to cut a 32" lead? 

10. What gears will be necessary to cut a 20 deg. spiral groove one 
complete turn on a piece 5.460" in circumference? 

Friction 

Friction is the resistance to motion which takes place 
between two bodies at their surface of contact, and depends 
upon the force with which the bodies are pressed together 
and upon lubrication. The force of friction will always act 



FRICTION 



131 



in a direction opposite to that in which the body tends to 
move. 

Coefficient of Friction is the ratio of the force required to 
slide a body along a horizontal plane surface to the weight 
of the body. 

When a body W is just on the point of moving by a force F, 
the amount of friction between the body and the table is 
called Static Friction. 

Example : A body weighing 30 lbs. ( W) rests on a horizontal 

surface. The force required to keep it in motion along the 

surface is 6 lbs. (F). Find the coefficient of friction. 

F 6 
Coefficient of friction'/ = — = — or 0.20 (Ans.). 

W 30 

The coefficient of friction is equal to the tangent of the 

angle of repose, which is the angle of inclination to the hori- 




F=6lbs 



zontal of an inclined plane on which the body will just over- 
come its tendency to slide. This angle is usually denoted 
by0. 

Coefficient of friction/ = tangent 6. 

f = 0.20 = tang. 6 or 11 deg. 19 min., or the angle at which 
the above weight would repose. 

A greater force is required to start a body from a state of 
rest than to merely keep it in motion, because the friction 
of rest is greater than the friction of motion. 

When a body rolls on a surface, the force resisting the 
motion is termed Rolling Friction, and has a different/ value 
than sliding friction. 



132 INDUSTRIAL MATHEMATICS 

Let W = total weight in lbs. of rolling body or load on 
"w wheels. 

R = radius of wheel, in feet. 




Force 

<^^i " / = coefficient of rolling friction. 

Then 

WXf 
resistance to rolling in lbs. = 

when force is applied radially, and parallel to plane. 

Table — Value of Coefficient of Friction (/) for Low Pressures 
(Sliding Friction) 

Bronze on bronze (dry 0.20 

Bronze on cast iron (dry) 0.21 

Cast iron on cast iron (lubricated) 0.15 

Cast iron on hard wood (dry) 0.49 

Cast iron on hard wood (lubricated) 0.19 

Leather on cast iron (dry) 0.56 

Hardwood on hard wood (dry) .0.48 

Steel on cast iron 0.20 

Steel on brass 0.15 

Steel on steel ..0.14 

Rolling Friction 

Iron on iron 0.003 

Wood on wood 0.005 

Iron on wood 0.018 

Rubber on asphalt 0.025 

EXERCISES 

1. What is the / between two bodies, if it requires a 50 lb. force to 
move a 175 lb. weight? 

2. What is the / between cast iron, if it requires a pull of 15 lbs. 
to slide a lathe carriage which weighs 80 lbs. upon its ways? 

3. What is the / of cast iron, if it requires a 26 lb. force to slide a 
145 lb. shaper ram upon its ways? 

4. What is the angle of repose of bronze on cast iron, if it requires a 
force of 1 lb. to slide a 48 lb. weight? 

5. What is the angle of repose for bronze on bronze? 

6. What is the angle of repose for hardwood on hardwood? 



ELECTRICITY 133 

7. What is the angle of repose for steel on cast iron?, 

8. If the angle of repose is 8 deg. 32 min. for cast iron on cast iron, 
what force will be required to pull a 500 lb. cast iron weight upon a 
cast iron surface plate? 

9. A flywheel weighing 400 lbs. is being rolled into a doorway 3^ ft. 
above the ground on a plank 16 ft. long. How much power must be 
applied parallel to the plank to keep the flywheel from rolling back? 

10. What force applied parallel with the floor, will be required to 
roll a 36" armature weighing 4125 lbs. along a floor, if /is equal to 0.012? 

Electricity 

A Volt is a unit of electrical pressure or potential difference 
(P.D.) or the electro-motive-force (e.m.f.) required to cause 
a current of one ampere to flow through a resistance of 
one ohm. 

An Ampere is a unit of current strength, or the quantity 
of flow, or the quantity of current which will flow through a 
resistance of one ohm under an electro-motive-force of one 
volt. 

An Ohm is a unit of resistance, or the resistance of a con- 
ductor through which a current of one ampere will pass under 
an electro-motive-force of one volt. 

A Watt is a unit of power or energy, 746 watts = 1 h.p. 
A watt is the amount of electrical energy being used when one 
ampere of current is flowing under a pressure of one volt. 

A Kilowatt is equal to 1000 watts. 

A Kilowatt-Hour is equal to a kilowatt of electrical energy 
continued for one hour or its equivalent. 

The resistance of conductors of the same material are 
directly in proportion to their length and inversely in pro- 
portion to their cross section. That is, doubling the length 
of a wire doubles its resistance, and doubling its diameter 
makes the resistance J as great, since it makes the cross 
section 4 times as great. 

1000 ft. of No. 10 B. & S. copper wire (0.102" diameter) 
has an approximate resistance of 1 ohm. 



134 INDUSTRIAL MATHEMATICS 

Formulas: 

E E 

C = — , E = C X R, R=—, E X C = watts, 

i? C 

E = electro-motive-force in volts. 
R = resistance in ohms. 
C = current in amperes. 

Problem I : How many amperes of current is flowing in a 
circuit with a resistance of 20 ohms when the pressure is 
ioo volts? 

E ioo 

C = — , C = — = 5 amperes (Ans.). 

i? 20 

Problem 2: What is the pressure when 5 amperes of cur- 
rent is flowing in a circuit in which the resistance is 20 ohms? 

E = C X R, E = 5 X 20 = 100 volts (Ans.). 

Problem 3: What is the resistance of a circuit when 5 
amperes of current flow under a pressure of 100 volts? 

E 100 

R = — , R = — = 20 ohms (Ans.). 
^ 5 

Problem 4: How many watts will be consumed, when a 
10 ampere current is flowing under a pressure of no volts? 
E X C = watts, 110X10= 1 100 watts (Ans.). 

EXERCISES 

1. (a) Find the resistance of 1 mile of No. 12 copper wire, if the 
resistance of 1 ft. of No. 12 wire is 0.002476 ohms. 

(b) What length of wire will have a resistance of 125 ohms? 

2. Find the resistance of 1000 ft. of No. 6 aluminum wire if 1 mile 
of this wire has a resistance of 3.3687 ohms. 

3. What is the resistance of 100 ft. of German silver wire, if 11" 
of this wire has a resistance of 0.022 ohms? 

4. What is the resistance hi ohms of a no volt line carrying a 10 
ampere current? 

5. Find the resistance of a 220 volt line carrying a 25 ampere current 



WORK, POWER AND THE STEAM ENGINE 1 35 

6. What current in amperes is carried on a line that has a resistance 
of 0.8 ohms on a no volt circuit? 

7. What ampereage is there in a 220 volt line which has a resistance 
of 50 ohms? 

8. What kw. motor can be run with a 50 ampere current on a 220 
volt line? 

9. What e.m.f. is there on a line that has a resistance of 200 ohms, 
when a 10 ampere current is flowing? 

10. What is the voltage of a battery that has a resistance of 0.25 
ohms and a 6 ampere flow? 

11. What e.m.f. exists between the ends of a wire whose resistance 
is 100 ohms, when the wire is carrying a current of 0.7 amperes? 

12. How much current will flow between two points whose P.D. is 
8 volts, if they are connected with a wire having a resistance of 350 
ohms? 

13. What will it cost per hour to operate a 5 ampere soldering iron 
on a no volt circuit. If 10 cents is being paid per kilowatt-hour. 

14. What will it cost per year to burn five 40 watt lamps on an average 
of 3 hours per day. If 6 cents is paid per kilowatt-hour. 

Horse Power Calculations, etc. 

Work. — The unit of work is the foot pound or the amount 
of work in overcoming a pressure or weight equal to one 
pound through one foot space, and the time in doing this is 
not considered. 

Formula: Weight in lbs. X distance in feet = ft. lbs. 

Example: If a man lifts a casting weighing 100 lbs. on a 
bench 3 ft. high, the work done is 300 ft. lbs. 100 X 3 
= 300 (Ans.). 

Power. — H.p. in mechanics is the power exerted or work 
done in lifting a weight of 33000 lbs. one foot per minute or 
550 lbs. one foot per second. Time to do the work is always 
considered in h.p. The power of an average horse is about 
f h.p. and that of an average man is about 1/7 h.p. Example: 
To lift a 3300 lb. casting 200 ft. high in 2 min., a 10 h.p. 
engine would be required. 



I36 INDUSTRIAL MATHEMATICS 

Formula: 

ft. lbs. 
h.p. = 



33000 X minutes 

3300 X 200 

Solution = == 10 (Ans.), 

33000 X 2 



The Steam Engine 

The power exerted by a piston driven by steam or other 
medium during the stroke in ft. lbs. is equal to the area of 
the piston times the pressure per square inch times the 
stroke in feet. In the case of steam engines, where the 
steam is cut off at J or ^ or J of the stroke, the piston, being 
driven the rest of the way by the expansion of the steam, 
the average pressure for the entire stroke is the mean effective 
pressure (m.e.p.) as it is called, is the basis of calculation. 
As each revolution of the engine equals two strokes of the 
piston, the number of foot pounds per minute times length 
of stroke in feet, times r.p.m. times 2 = a product which 
divided by 33000 equals indicated h.p. (i.h.p.) or h.p. de- 
veloped in cylinder. Example: An engine with a 10" bore 
and a 12" stroke running at a 100 r.p.m. with a m.e.p. of 
80 lbs. will develop 38 plus h.p. 

Formula 1 : 

area X m.e.p. X stroke X r.p.m. X 2 PLAN 

i.h.p. = ■ = X 2. 

33000 33000 

Solution: 

10 X 10 X 0.7854 X 80 X 1 X 100 X 2 ' /a x 

= 38 + h.p. (Ans.). 

33000 

Formula: 

i.h.p. X 33000 



Area of piston = 
Area X stroke = 



m.e.p X stroke X r.p.m. X 2 
i.h.p. X 33000 



m.e.p. X r.p.m. X 2 



m.e.p. 



THE STEAM ENGINE 1 37 

i.h.p. X 33000 



area X stroke X r.p.m. X 2 

Gas Engine H.P. Formulas 

Brake h.p. = power actually delivered or i.h.p. minus 
friction of engine. 

Formula 2: A.L.A.M. rating 

U l X N 
h.p. = . 

2-5 

D = Diameter of cylinder in inches. 
N = No. of cylinders. 

This formula is taken as a standard for 4 cycle single- 
acting engines at a piston speed of 1000 feet per minute. 

Example: A 4 cylinder engine with 4§" bore will develop 
32i h.p. 

Formula 3: 

D*XPXLXRXN 

i.h.p. = . 

1,000,000 

D = Diameter or bore of cylinder. 

P = m.e.p. average about 80 lbs. 

R = r.p.m. 

N — No. of cylinders. 

L = Length of stroke in inches. 

Example: What is the i.h.p. of a 4 cyl. engine 4§" bore 
and sf " stroke running at 1000 r.p.m. with a m.e.p. of 80 lbs.? 



Solution: 

20.25 X 80 X 5-75 X 1000 X 4 
1,000,000 



= 37.26 i.h.p. (Ans.), 



I38 INDUSTRIAL MATHEMATICS 

EXERCISES 

1. How much work in ft. lbs. will a man do per 10 hr. day, if he 
places 3800 bricks per hour on a wagon 2\ ft. high, each brick weighing 
86 oz.? 

2. What force in lbs. does a man exert if he lifts a 1750 lb. casting 
8 ft. high with a crane having a velocity ratio of 10 to 1? 

3. What horse power does this man exert (above problem), providing 
he lifts the casting in 180 seconds? 

4. A locomotive having two cylinders with a 26" bore, 36" stroke, 
running at 60 r.p.m. with a m.e.p. of 90 lbs. will develop what hp.? 

5. How much work must be done to raise 120 long tons of coal from 
a mine 216 ft. deep? What must be the hp. of an engine to do it in 
jour hrs. if the friction of the machinery increases the work 10 percent? 

6. A cylindrical well 4 ft. in diameter and 72 ft. deep has 16 ft. of 
water in it. What must be the h.p. of the engine to empty this well in 
40 minutes? 

7. What must be the bore of an engine to develop 25 h.p. running 
275 r.p.m. with a 12" stroke and a m.e.p. of 60 lbs.? 

8. What is the A.L.A.M. rating of a 2 cylinder gas engine having a 
4" bore? 

9. What is the i.h.p. of a 4 cylinder gas engine running at 1200 
r.p.m. having a z\" bore and z\" stroke, with a m.e.p. of 90 lbs. (using 
formula No. 3). 

10. What h.p. will a 4 cylinder motor cycle develop running at 1400 
r.p.m., with a 2" bore and 3" stroke, m.e.p. 80 lbs.? (Using formula 
No. 2 and No. 3.) 

Strength and Proportion of Gear Teeth 

The strength of gear teeth and the h.p. that may be 
transmitted by them depends upon so many variable and 
uncertain factors that sometimes it involves rather compli- 
cated formulas. 

The various elements which enter in the constitution of a 
formula to represent the working strength are as follows: 

1. The strength of material, which is often a variable quan- 

tity. 

2. The shape of tooth, which is sometimes under-cut. 

3. The point of application of the load (pitch line or extreme 

end). Some authorities differ on this subject. 



STRENGTH AND PROPORTIONS OF GEAR TEETH 1 39 

4. Whether we consider the load applied to one or more 

teeth. 

5. Pitch line velocity. 

6. Factor of safety. 

Formula 1 — for cast iron gears: 

0.910 X V X P X F 
h.p. = -. . 

Vi +0.65F 

V = velocity of pitch line in ft. per sec. 
P = circular pitch in inches. 
F = face of gear in inches. 

The width of face is generally 2 to 3 times the circular pitch. 
Formula 2: 

pitch diameter X r.p.m. X C.P. X face X 200 



h.p. = 



C.P. X face = 



26050 
[26050 X h.p. 



P.D. X r.p.m. X 200 



EXERCISES 

1. What h.p. will the following gear safely transmit: 24" P.D., 
3.1416 C.P. 4" face, running at 100 r-p.m.? 

2. What h.p. can be transmitted with a cast iron gear 2\" P.D., 
10 P, 1" face, running at 500 r.p.m.? 

3. What pitch gear must be used to transmit 1 H.P. with a cast 
iron gear 3. 151" P.D., running at 200 r.p.m., with a 1" face? 

4. What size face will be required on a gear 8" P.D., 3 P, running 
at 350 r.p.m., to transmit 20 h.p.? 

5. What pitch gear will be required on a hand crane, to be able to 
lift a 66000 lb. casting 10 ft. high in 10 minutes, the pinion having a 2" 
face, 8" P.D., being turned by a crank at the rate of 50 r.p.m.? 

6. What must be the width of the face of a pinion 10" P.D., 1 P, 
on a steam engine, running at 200 r.p.m., driving a derrick that is 
capable of lifting a 10 ton I beam 33 ft. in 1 min.? 

7. What h.p. will a gear of 10 P, 2" P.D., 13" face, transmit running 
at 600 r.p.m.? 

8. What h.p. can be safely transmitted with a 6 P gear, 6" P.D., 
with a 1" face, running at 250 r.p.m.? 



140 



INDUSTRIAL MATHEMATICS 



9. What width face will be required on a gear that is 20" P.D., 2 P, 
running at 200 r.p.m., to transmit the maximum strain produced by a 
steam engine running at 275 r.p.m. with a 12" stroke, 8" bore and 
m.e.p. 120 lbs.? 

10. A 10 h.p. engine driving a stone crusher through spur gears 
running at 150 r.p.m., having 10" P.D., gears and 2" face, will require 
approximately what pitch gears? 

Resolution of Forces 

Any single force which produces the same effect on a body 
as the combined action of two or more forces is called the 
Resultant of those forces. The resultant of two or more 
forces can be found either graphically or algebraically. 

The resolution of two forces acting in the same straight 
line, in the same direction, is the sum of the given forces. 
Example: A force of 5 lbs. and 7 lbs. acting in an easterly 
direction on a body (A) will create a force of 12 lbs. in the 
easterly direction. 



< — 5 lbs-.— > 




<- 7 lbs.-- -> 


A 







•7 lbs. -><— -5 Ibsr— > 



Graphical Representation. 



k- 12 lbs. -* 

Resultant Force. 



Fig. I 



The resultant of two forces acting in the same straight 
line, but in opposite directions, is the difference of the given 
forces and act in the direction of the greater. Example: If 
there is a force of 5 lbs. on an object (^4) in a westerly direc- 
tion and a force of 7 lbs. acting in an easterly direction, the 
resultant force would be 2 lbs. in the easterly direction. 



<- — 


—-5 lbs.— 


— -> 






7 IbSr -> 


A 













Graphical Representation. 





*-2lbs-.->| 


A 


i 


-* Y 



Resultant Force. 



Fig. II 



RESOLUTION OF FORCES 



141 



The resultant of any two forces acting at an angle to each 
other may be found by completing the parallelogram upon 
the forces as sides, and drawing the diagonal. Example: 
Suppose a force of 6 lbs. is acting on an object, (A) in an 
easterly direction, and a force of 8 lbs. in the southerly 
direction, then the resulting force will be equal to the diagonal, 
or 10 lbs., in the southeasterly direction. 



8 lbs. 




Graphical Representation. 
Fig. Ill 




Fig. IV 



Suppose a crane (Fig. IV) lifting a 1000 lb. weight: what 

would be the compression in lbs. on member B, and what 

would be the tension on tie rod A, if angle C was 60 deg. 

S.O. = 1000 sine of 30 deg. = 0.5000. Hyp. = S.O./sine 

or 2000 lbs. = compression on member B. S.O. X cotang. 

= S.A. or 1000 X 1. 7321 = 1732. 1 lbs. = tension on tie 

rod A. 

EXERCISES 

1. If it requires two locomotives, one of 700 h.p. and the other 800 
h.p. to pull a freight train, what is the total power in ft. lbs. per minute? 

2. If there are 10 men, each pulling with a force of 60 lbs. on one 
end of a rope, and 12 men on the other end pulling with a 50 lb. force, 
what is the tension on the rope in ft. lbs., arid what will be the direction 
of the resultant force? 

3. Two boys are rowing a boat in the middle of a river 1 mile wide, 
running north and south. "A" rows east at the rate of 3 miles per 
hour. "B" rows west at the rate of 4j miles per hour. Which boy 
and how long will it take him to reach shore? 

4. If in Fig. IV, a crane is lifting an 8 ton weight, and the angle 
C is 50 deg., what will be the tension on A and compression on B. 

11 





142 INDUSTRIAL MATHEMATICS 

5. Three ropes are fastened to a ring, and A 
\A° 5 lbs, is pulling on a rope with a force of 50 lbs. east. B 
is pulling on a rope with a force of 50 lbs. south. 
1 50 lbs ^ wnat an gl e an d with what force will a man have to 

pull on the third rope to keep the ring stationary. 

6. A force of 20 lbs. is acting vertically upward and is resolved into 
two forces, one of which is horizontal and equal to 10 lbs. What is the 
direction and magnitude of the component? 

7. If a casting on a truck requires a force of 1000 
lbs. to move it, what two equal forces will be required 
to move it if the forces are acting at right angles to 
each other? 

8. What will be the direction and magnitude of a 
force to equalize a force of 100 lbs. in an easterly direc- 
tion and a force of 200 lbs. in a southerly direction? 

9. What will be the direction and magnitude of a force to equalize 
a force of two tons in a southerly direction and a 3500 lb. force in an 
easterly direction? 

10. What will be the pulling strain on a rope if a boy pulls with a 
force of 50 lbs. on one end, and two boys pulling in the opposite direction 
with a force of 25 lbs. each? 

Falling Bodies 

Gravity attracts all bodies toward the center of the earth 
with an acceleration which varies with the location of the 
body relative to the distance from the center of the earth. 
Theoretically acceleration is the same on all bodies, inde- 
pendent of their size, weight or shape. 

A freely falling body, or a body moving under the influence 
of gravity (friction not considered), at the end of the first 
second, will have a velocity of 32.16 ft. per second. During 
the next second, it will acquire 32.16 ft. additional velocity, 
giving it a velocity of 64.32 ft. at the end of the second 
second. Each succeeding second will add 32.16 ft. to the. 
velocity the body had at the end of the preceding second. 
This acceleration as it is called, due to gravity or 32.16 ft. 
in formulas, is designated by the letter "g." 



FALLING BODIES 1 43 

V = velocity in feet per second. 
g = acceleration due to gravity. 
T = time in seconds. 
5 = space in feet. 
H = height in feet. 

To find the Velocity (V) of a falling body at the end of 
any number of seconds, multiply 32.16 by the number of 
seconds the body has fallen. Example: Find the velocity 
of a falling body at the end of 10 seconds. 

Formula: V = G X T. V = 32.16 X 10, or 321.6 ft. 
(Ans.) or V = 25/ T. 

The distance (S) through which a body falls in a given 
time equals the square of the number of seconds during 
which the body has fallen, multiplied by 16.08 or \g. Total 
distance fallen = S. Example: Find the total distance a 
body will fall in 10 sec. 

Formula: S = T 2 X \g = io 2 X 16.08 = 1608 ft. (Ans.) 
or S = iVT or S = V 2 J2g. 

The time in seconds ( T) required for a body to fall a given 
distance equals the square root of the distance expressed in 
feet, divided by 4.01. Example: Assume that a stone falls 
through a distance of 1608 ft. How long will it take to 
drop? . 

Formula: 



„ Vs V1608 /A . ^ v „ 2s 

1 = = = 10 sec. (Ans.) or T = — or T = — . 

4.01 4.01 g V 

The velocity ( V) of a falling body after it has fallen through 
a given distance equals the square root of the distance through 
which it has fallen, times 8.02. 

Example: What is the velocity of a falling body after it 
has fallen through a distance of 1608 ft.? 

Formula: V = 8.02 X VS. 8.02 X V1608 = 321.6 ft. 
per sec. (Ans.) or V = ->J2gS. 



144 INDUSTRIAL MATHEMATICS 

The height (H) through which a body must fall to acquire 

a given velocity equals the square of the velocity divided 

by 64.32 or (2 X G). Example: From what height must a 

body fall to acquire a velocity of 321.6 ft. per sec? 

V 2 32 1. 6 2 

Formula: H = — = = 1608 ft. (Ans.). 

2g 64.32 

If a body is thrown vertically upward with a given velocity, 

its velocity will be retarded during each second in the same 

ratio as it is accelerated when falling. A body thrown 

vertically upward in the air will return to the ground with 

exactly the same velocity as that which it had when thrown 

up. At any point the velocity going up will be equal to the 

velocity going down except that the direction is reversed. 

EXERCISES 

1. What velocity has a body obtained after having fallen 5 seconds? 

2. How far will a body fall in 15 seconds? — in 30 seconds? 

3. How many seconds will it take for a body to fall 1000 ft.? 

4. A body has fallen 100 ft. Find its velocity in ft. per sec. at the 
last sec. 

5. How far must a body fall to acquire a velocity of 50 ft. per 
sec? — 100 ft. per sec? 

6. A person falls from a 22 story building which is 200 ft. high. 
What will be the velocity of the body? How long will it take the body 
to drop? 

7. An object is thrown vertically upward with a velocity of 100 ft. 
per sec When does it reach the highest point and start to return? 
How high will it go? 

8. A body falls from a balloon to the earth in 10 seconds. What is 
the height of the balloon? What is the velocity of the falling body? 

9. A falling body passes a given point (A) at a velocity of 15 ft. 
per second. How far below the point (A) is the body after 5 seconds? 
How far does the body fall during the fifth second after passing the 
given point 04.)? 

10. With what velocity must a ball be shot upward to rise to the 
height of the Washington Monument, which is 555 ft. high? 



CENTRIFUGAL FORCE 1 45 

Centrifugal Force. 

When a body revolves in a curved path, it exerts an out- 
ward force called Centrifugal force upon the arm or cords 
which restrains it from moving in a straight line. Example 
of this action, mud flying from a carriage wheel, bursting of 
emery wheels and fly wheels. This force depends upon the 
mass, velocity and radius of the revolving mass. 

F = centrifugal force. R = radius in feet. 

W = weight or mass. N = r.p.m. 

V = velocity in feet per sec. G = gravity or 32.16. 
Formulas: 



F = 


WV 2 
GR ' 


WV 2 

R = , 

FG 


N - J 29i3F 
v WR 




W = 


FRG 


Ifrg 



Example: Find the centrifugal force or tension on the 
spokes of a fly wheel which is 6 ft. in diameter, running at 
60 r.p.m., providing the rim weighs 300 lbs. 

Formula: 

WV 2 300 X (1 X 6 X tt) 2 

F = — = " K -— " = 1104.8 lbs. (Ans.). 

GR 32.16 X 3 

For Thin Disks such as an emery wheel, rotating about its 
center, the sum of all radial or centrifugal forces that tend 
to rupture the disk equals 0.00000835 WRN 2 . 

EXERCISES 

1. What is the centrifugal force on a fly wheel 10 ft. in diameter 
running 60 r.p.m., providing the rim weighs 500 lbs.? 

2. A body weighing 200 lbs. is moving with a velocity of 40 ft. per 
sec. in a circle 10 ft. in diameter. What pressure is required to keep it 
in its path? 

3. What will be the unbalanced strain on an automobile wheel, if 
there is a 1 lb. air valve on the rim 18" from center, and the wheel 
running at 500 r.p.m. (about 55 miles per hour)? 



I46 INDUSTRIAL MATHEMATICS 

4. What will be the strain on an emery wheel 24" in diameter, 
weighing 50 lbs. running at 4000 surface ft. per minute? 

5. What will be the maximum weight of a rim of a fly wheel, it 
being 4 ft. in diameter, running at 60 r.p.m., and the spokes are of such 
size that they can resist a strain of 600 lbs.? 

6. What will be the maximum speed in ft. per min. that a fly wheel 
6 ft. in diameter can be run, the rim weighing 400 lbs., with spokes 
of the same strength as in the above problem? 

7. What diameter can we make a fly wheel within a safe limit, 
running at a surface speed of 3600 ft. per min., the rim to weigh 600 lbs. 
and the spokes to resist a strain of 2000 lbs.? 

8. A locomotive weighing 100 tons, is running at a speed of 15 miles 
per hour. What side pressure will the fail receive when passing around 
a durve 120 ft. radius? 

9. What will be the strain on an emery wheel 6" in diameter, running 
at a surface speed of 5000 ft. per min., the wheel weighing 36 ounces? 

10. A 10 lb. weight fastened on the end of a cord 5 ft. long is revolved 
at the rate of 100 r.p.m. What is the tension on the cord? 

Horse Power of Belting 

The h.p. a belt can safely transmit depends principally 
upon the velocity of the belt, the working stress or pull in 
pounds per inch of width the coefficient of friction and the 
arc of contact. The most economical speed for belting is 
between 4000 and 4500 ft. per min. In higher speeds than 
this, the action of centrifugal force is a hindrance. The most 
satisfactory working stress is about 575 to 850 lbs. per sq. 
in. of belt section. A commonly used value for the effective 
pull of a single belt is 35 lbs. per in. width, and 60 lbs. for 
double belt. A single belt is about 3/16" thick and weighs 
approximately 16 oz. per sq. ft. A double belt is twice the 
thickness and weight. The ultimate strength of oak tanned 
leather runs from 3000 to 7000 lbs. per sq. in. 

Rule 1. — A single belt 1" wide running at 1000 ft. per min. 
will transmit 1 h.p. 

Formula 1: 

PXDXNXW „ r h.p. X 33000 X 4 

h.p. = , W = , 

4 X 33000 P XD X N 



HORSE POWER OF BELTING 147 

Formula 2: 

PVW h.p.X 33000 

h.p. = , W = — . 

33000 PV 

D — Diameter of driving pulley in inches. 

V = Velocity of belt in ft. per min. 

N = r.p.m. of driver. 

P = Effective pull of belt per inch of width in lbs. 

W = Width of belt in inches. 

EXERCISES 

1. What h.p. will a 6" single belt transmit running at 4000 ft. per 
minute, using rule 1? 

2. What width belt must be used to transmit 25 h.p. if the belt is 
running at 4500 ft. per minute, using rule 1? 

3. What h.p. will a single belt 10" wide transmit if it is running 
over two 10 ft. pulleys, running no r.p.m. What will a double belt 
transmit? 

4. What width double belt must be used to safely transmit 50 h.p., 
if it is running at 4500 surface ft. per min., using formula 2? 

5. What width single belt must be used to transmit 30 h.p. if the 
driver is 4 ft. in diameter, running at 400 r.p.m.? 

6. A 10" double belt driven by a 3 ft. driver, running at 300 r.p.m. 
will transmit what h.p.? 

7. Using rule 1, what h.p. will a 5" single belt transmit, running at 
4500 surface ft. per minute? 

8. According to rule 1, what width belt must be used to transmit 
'8 h.p. if the belt is running at 5000 surface ft. per minute? 

9. A 10 h.p. stationary gas engine, running at 400 r.p.m., has a 
driving pulley 10" in diameter. What must be the width of a double 
belt to safely transmit its full power? 

10. What must be the width of a single belt to safely transmit the 
power of a i\" h.p. motor, having a driving pulley 8" in diameter 
running at 1500 r.p.m.? 

Length of Belting 

When Both Pulleys are of the Same Size. — Multiply the 
diameter of either one of the pulleys by tt and add to this 
twice the distance between the shafts. 



I48 INDUSTRIAL MATHEMATICS 

When One Pulley is Considerably Larger than the Other. — 

Square the distance between centers of the shafts, and add 
to this the square of the difference between the radii of the 
two pulleys. From this total, extract the square root, and 
multiply by 2 (calling this A). Add the diameter of the 
two pulleys and multiply by 1.57. Add this result to A, 
which will be the total length of belt. 

Formula: 
Length = [{ VCD. 2 + (R - r) 2 } X 2] + [_{D + d) X 1.57J 

For Cross Belts. — 

(^4) Square the diameter of the large pulley and the dis- 
tance between centers, add together and extract the square 
root. 

(B) Square the diameter of the small pulley and the 
distance between centers, add together and extract the 
square root. 

( C) To the sum of the two square roots, add the product 
of the sum of 2 pulley diameters X 1.57. The total will be 
the required length. 

Formula: 



Length = [ V(£> 2 + CD. 2 ) + V(d 2 + CD. 2 )] 

+ [_{D + d) X 1.57]. 
EXERCISES 

1. What will be the length of the cross belt required to go over two 
pulleys 2 ft. and 4 ft. in diameter, shafts being 25 ft. apart? 

2. What will be the length of belt to use over pulleys, both being 
3 ft. in diameter, shafts 12 ft. apart? (For cross belt.) 

3. Two pulleys 52" and 24" in diam., are on separate shafts 15 ft. 
between centers. What will be the length of belt required? 

4. What is the length of cross belt running over two pulleys 16" 
and 24" in diameter, shafts being 20 ft. apart? 

5. What is the length of cross belt 18 ft. between center of pulleys 
that are 8 ft. and 3 ft. in diameter? 

6. What will be the length of belt required, if the center distance 
between shafts is 15 ft., the driver being 8" in diameter, running at 
1500 r.p.m. and the driven at 275 r.p.m.? 



ROPE DRIVES 149 

7. A gas engine has a drive pulley 10" in diameter, running at 900 
r.p.m., driving a line shaft 15 ft. away at 500 r.p.m. What will be the 
length of cross belt required? 

8. Two pulleys 2 ft. and 5 ft. in diameter, on separate shafts 12 ft. 
apart will require what length of belt? 

9. What will be the length of cross belt required for the above 
problem? 

10. What is the length of cross belt, shafts being 30 ft. between 
centers, pulleys being 60" and 48" in diameter? 

Rope Drives 

Rope Drives are used for long distant transmission, and 
the effective pull of the rope depends upon the friction in the 
V shaped grooves on the pulley. These grooves are between 
45 and 60 degrees. 

Rope Pulley Diameters. — The diameter of the pulleys 
should not be smaller than 30 times the diameter of the 
rope to prevent internal wear, due to the fibers sliding upon 
each other. 

Speed for Rope Transmission. — The rope should be run 
at a speed between 4000 and 5500 surface feet per minute. 
If they are run faster, the action of the centrifugal force is a 
hindrance. The tensile strength of manila rope in some cases 
is as high as 50,000 lbs. per sq. in. The strength of cotton 
rope is about four-sevenths of that of manila rope. 

Center Distance depends upon the size of the pulleys and 
the size of the rope. This should not be less than 25 feet 
and not over 300 feet. If a longer drive is necessary, a 
series of drives may be arranged, or else the rope must be 
supported by loose pulleys to prevent excessive sag. 

Ropes from 1" to if " in diameter are generally used. The 
safe working stress for rope is about 200 lbs. per square inch 
of sectional area. 

Formula: D 2 X 0.3 = approx. weight in lbs. per linear ft. 
of rope. 



150 INDUSTRIAL MATHEMATICS 

Formula: h.p. = D 2 X V X 0.003 X N. 

D = diameter of rope in inches. 
V = velocity in ft. per minute. 
N = no. of ropes. 

„ / h.p. h.p. 

^ = Vr ■■■■ N 



V X 0.003 X N ' D 2 X V X 0.003 ' 

EXERCISES 

1. What h.p. will a 1" rope transmit, running at 5000 ft. per minute? 

2. What h.p. will a 2" rope transmit, running at 5000 ft. per minute? 

3. Find the weight of 500 ft. of i|" rope. 

4. What is the weight per ft. of a i\" rope? — of a 3" rope? 

5. An engine having a 20 ft. drive wheel, running at 80 r.p.m., 
driving a line shaft with 10 — \\" ropes will transmit what h.p.? 

6. How many i\" ropes must be used to transmit 100 h.p., if the 
surface speed is 4500 ft. per minute. 

7. How many 1" ropes are required to transmit 175 h.p. if the drive 
pulley is 10 ft. in diameter, running at 150 r.p.m.? 

8. What will be the total weight of rope in the above problem if 
each rope is 125 ft. long? 

9. I have a 50 kw.. motor running at 500 r.p.m. with a 3 ft. 
drive pulley. How many 1" ropes should I use? 

io. What size rope must be used to safely transmit the full power 
produced by a Corliss engine, having a 36" stroke, 24" bore, running 
at 150 r.p.m. with a m.e.p. of 125 lbs., providing 10 ropes are to be 
used on a drive wheel 10 ft. in diameter? 

Cable or Wire Rope Drives 

Wire Rope Drives are used for long distance transmission 
or where atmospheric conditions will not permit other 
methods. The effective pull of the wire rope depends upon 
the friction at the bottom of the grooves, the radius of 
which should be slightly larger than that of the rope. These 
grooves, are generally lined with leather, wood, rubber, or 
some other similar material to increase the friction between 
the rope and sheave and at the same time to reduce the 
wear of the rope to a minimum. 



CABLE OR WIRE ROPE DRIVES 151 

Cable Pulley Diameters should not be less than seventy 
five times the diameter of the cable. 

Speeds for Cable Transmission should be between 4000 
and 5000 surface feet per minute and the lower rope should 
be the one under tension to increase the arc contact. 

Center Distance depends upon the size of pulleys and the 
size of cable. It should be between 60 and 400 feet. If a 
longer drive is necessary, the cable must be supported by 
loose pulleys to prevent excessive sag or a series of drives 
may be arranged. 

Formula: D 2 X 1.58 = approx. weight in lbs. per linear 
ft. of cable. 

Formula: h.p. = D 2 X V X 0.060 X N for iron wire. For 
steel wire, multiply by 2. 



No. of cables = h ' P - ' h ' p - 



*>-4- 



D 2 X V X 0.060 \FX 0.060 X N 

EXERCISES 

1. What h.p. will a 1" iron cable transmit, running at 5000 ft. per 
minute? 

2. What h.p. will two f" steel cables transmit, running at 4000 ft. 
per minute? 

3. Find the weight per ft. of a 1" cable. Of a i§" cable. 

4. What is the weight of 700 ft. of f" cable? 

5. How many 1" iron cables must be used to transmit 125 h.p. ( 
if the surface speed is 5000 ft. per minute? 

6. What h.p. can be transmitted by a \" iron cable, at 100 r.p.m. 
on a 5 ft. diameter drive sheave? 

7. I have a 200 kw. motor running at 200 r.p.m., with an 8 ft. drive 
wheel. How many 1" cables should I use? 

8. How many §" cables are required to transmit 200 h.p. if the 
surface speed is 4000 ft. per minute? 

9. What size iron cables must be used to transmit the full power 
developed by a cross compound Corliss engine, having a 4 ft. stroke. 
The high pressure cylinder has a bore of 24" and a m.e.p. of 80 lbs., 
and the low pressure cylinder has a 44" bore and a m.e.p. of 25 lbs. 
Two ropes are used on a drive wheel 20 ft. in diameter, running at 65 r.p.m. 



152 INDUSTRIAL MATHEMATICS 

10. I have an 80 h.p. gas engine running at 300 r.p.m., with a 5 ft. 
drive pulley, driving a line shaft 50 ft. away at 200 r.p.m. What size 
steel cable must be used and what will be the weight of the cable? 

Chain Transmission 

Chain Drives are used for positive transmission of very 
heavy powers at rather slow speeds. There are several 
different types of chains, the round link chain used consider- 
ably on agricultural machines, the flat link and block chain 
as used on bicycles, and the silent chain. Chains should be 
well lubricated, or if possible run in oil. 

Size of Sprocket Wheel. — The number of teeth in the 
sprocket should be as large as is consistent with other con- 
ditions. The best conditions are obtained with 16 or more 
teeth, although as few as 10 teeth can be used for slow 
speeds. The bottom diameter should be as accurate as 
possible, for on this diameter the chain rests. To obtain 
best results sprockets should always be machine cut. 

Speed of Chain Drives should not exceed 2000 ft. per 
minute. Best conditions are obtained with a speed of 1000 
ft. or under. 

Center Distance should not be less than ij times the 
diameter of the large sprocket, nor more than 12 ft. If 
longer drives are necessary, a 
series of drives should be ar- 
ranged; or else the chain must 
be supported by loose pulleys to 
prevent excessive sag, which 
causes the chain to whip. The 
center distance should be adjust- 
able if possible, to regulate the 

. c FlG - l 

amount of sag. 

The Safe Working Load of a chain depends on the amount 
of rivet bearing surface, and varies from 1/10 to 1/40 of the 
tensile strength, according to the speed, size of sprocket, etc. 




CHAIN TRANSMISSION 

Chain Sprocket Diameters, etc. 



A — -» 



153 




Tan C 





Fig. II 
P = Pitch. 

N = Number of teeth. 
b = Diameter of round part of chain block 
(.325" for 1" P, and .532" for i|" P). 
B = Center to center of holes in chain block 
e.400" for 1" P, and .564" for i|" P). 
A = Center to center of holes in side bars 

(.600" for 1" P, and .936" for i§" P). 
180 
N _. . .. _A_ 

Sin C 



Sin 



B 180 

— + Cos -— 
A N 



Pitch diameter 



O.D. 



P.D. + b. 



B.D. = P.D. 



To Calculate Length of Chain 

As a chain cannot contain a fractional part of a pitch, the 
next whole number above the calculated number of pitches 
is generally used. The chain length in inches is found by 
multiplying the number of pitches by the pitch in inches. 



EXERCISES 

1. What is the P.D. of a 10 tooth sprocket wheel with a i' 
chain? 



pitch 



154 



INDUSTRIAL MATHEMATICS 



2. What is the O.D. and B.D. of an 18 tooth sprocket with a i' 
pitch chain? 




Fig. Ill 
P = Pitch of chain. 
C = Center distance in pitches. 
N = Number of teeth on large sprocket. 
n = Number of teeth on small sprocket. 



L = Chain length in pitches 



• N n o.Q257(iV 
2C -\ — 

2 2 C 



n)* 



3. What is the distance A-B and diameter b of a if" pitch chain? 
(using the same proportions as a 1" pitch chain). 

4. Find the P.D., O.D. and B.D. of a 30 tooth sprocket for a \\" 
pitch chain. 

5. Find the B.D. of a 16 tooth sprocket for a x\" pitch chain. 



Shaft Design 

Shafting is generally made of cold rolled steel (C.R.S.), 
and is subject to two stresses: Shear stress due to the tension 
of the belt and weight of the shaft and pulleys: and twisting 
or torsional stress produced due to the resistance of the load. 

Shafting is generally calculated for torsional stress only 
although for light shafting with heavy pulleys, etc., bending 
stresses must be taken into consideration. 

Pulleys should be fastened as close as possible to the bearing 
to prevent excessive deflection. 

- The Allowable Stress for shafting varies from 5000 to 9000 
lbs. per sq. in. 



SHAFT DESIGN 155 



H.p. formulas: 
For long main power transmission shafts 



D*N 3/80 h.p. 



h.p. = . Diameter 



80 \ N 

For regular line shafts carrying drive pulleys 



h.p. = . Diameter = \/ 

54 ^ N 

For small short shafts well-supported 



D*N 3/40 h.p. 
h.p. = . Diameter = \ 

40 \ N 

Angle of torsional deflection for round shafts: 

583.6TL 



A = 



D*E 



The torsional deflection of round shafts is 70% greater than 

that of a square one. 

A = angle of deflection in degrees. 

D = diameter of shaft in inches. 

E = torsional modules of elasticity about 12,000,000 for 

steel shafting. 
L = length of shaft in inches. 
N = r.p.m. 
T = twisting moment in inch pounds. 

Torsional deflection of shafting should not exceed 5 min., 
or about 0.08 deg. per linear ft. 



4.6 



4/h.p. 



Linear deflection or sag should not exceed 0.012" per linear 
ft. The maximum distance in feet between bearings in 
order to prevent excessive sag should be calculated according 
to the following formula: 

L = a/i4o£> 2 . 



156 



INDUSTRIAL MATHEMATICS 



EXERCISES 

i. Find the torsional deflection of a 3" shaft 6 ft. long, subject to a 
twisting moment of 25000 in. lbs. 

2. What should be the diameter of a line shaft to transmit 7^ h.p., 
shaft running at 275 r.p.m.? 

3. What h.p. will a 2" shaft transmit, shaft running at 350 r.p.m.? 
(using formula for short shafts). 

4. Find the diameter of a line shaft to transmit 50 h.p. at 100 
r.p.m., with a torsional deflection not to exceed 0.08 deg. per linear ft. 

5. What should be the maximum length between bearings for a 1" 
shaft?— for a 2" shaft? 

6. Find the h.p. that can be transmitted by a \\" shaft, at 150 
r.p.m.? (using formula for short shafts). 

7. What diameter shaft is required to transmit 700 h.p., running at 
125 r.p.m.? 

8. Find the torsional deflection of an 8" shaft 10 ft. long, subject 
to a twisting moment of 20,000 in, lbs. 

9. What should be the maximum distance between bearings for a 
3" shaft? 

10. Find the diameter of line shaft to transmit 500 h.p. at 200 r.p.m., 
with a maximum torsional deflection of 0.07 deg. per linear ft. 



•OiU 



Scraped 
'Bearings 




Bearing Design 

The Size of Bearing required depends upon the bearing 
material, quality of oil used, speed, bearing pressure running 
conditions, etc. 



BEARING DESIGN 



157 




Fig. 



The Projected Area of a Journal 

(Fig. I) is the size used in calculation 
of the bearing surface; thus, the area 
of bearing surface for a 3" X 10" 
journal is 3" X 10" or 30 sq. in. 

The Amount of Pressure (P) al- 
lowed per sq. in. of projected area of 
bearings varies from 400 to 2000 lbs. 

The Diameter of shaft or pin must be designed strong 
and rigid enough to carry the required load. To do this 
the approximate length must be known. 

The Length of Bearing must be designed to have the 
proper bearing surface so that the unit pressure shall not 
exceed the allowable load. In general practice the length of 
bearing is from one to three times the shaft diameter. 

Formulas: 

A = allowable pressure per sq. in. of projected area of 

bearing. 
D = diameter of bearing in inches. 

drop feed 800, 



Y = quantity depending on 
method of oiling 

L = length of bearing in inches 
N — r.p.m. 



force feed or ring oil 1400, 
best conditions 2000. 



P = maximum safe unit 
pressure 



400 to 600 for shaft bearings, 

800 to 1000 for car journal, 

1200 to 1400 for crank pins, 

1600 to 2000 for wrist pins 

and sliding shoes. 

W = total load upon bearings in lbs. 

PY 

A = . 

(DN) + Y 

For bearings in the form of a sliding shoes or ways, the 
12 



158 INDUSTRIAL MATHEMATICS 

quantity 250 V is substituted for the quantity DN in the 
formula. V is the velocity of the rubbing surface in ft. 
per sec. To find the proper length of bearing use the follow- 
ing formula: 



W ( Y\ 



If the above formula gives a journal too long for practical 
use, the proper proportion of diameter and length should be 
adjusted to meet the required conditions. 

If the required projected area is known, the following 
formula may be used to get the proper proportion of the 
length to the diameter: L : D : : (§ ViV) : 1. 

EXERCISES 

1. What will be the allowable pressure per sq. in. on a car journal, 
using drop feed lubrication, if the journal is 4" in diameter, running at 
80 r.p.m.? 

2. What length bearing will be required in the above problem if 
there are 8 such bearings under a box car which weighs 18 tons and 
has a carrying capacity of 80,000 lbs.? 

3. Design a main bearing and an outboard bearing for an engine, 
providing a 16" shaft is to be used. The weight of the flywheel, shaft, 
crank-pin and ^ of connecting rod and other movable parts supported 
by bearings weigh 100,000 lbs. and 3/5 of this weight comes on the 
main bearing, and the remainder on the outboard bearing, the engine 
running at 100 r.p.m. (Ring oiling system to be used.) 

4. What length of crank pin bearing must be used for an engine with 
a 10" bore, running at 100 r.p.m., maximum steam pressure of 125 lbs. 
and force feed lubrication, if the crank pin is 2-J" in diameter? 

5. The 2ys" diameter line shaft in Dept. T-20 is 62 ft. long, with 
bearings 10 ft. apart, total weight of 27 pulleys and belt strain being 
approximately 3360 lbs. What should be the length of the journals 
used, if the shaft turns 280 r.p.m.? 

6. What should be the length of 2 journals on an electric motor 
running at 1190 r.p.m., the armature, commutator, pulley, belt strain 
and shaft weigh 1-500 lbs., providing the shaft is 2" in diameter, using 
ring oil lubrication? 



BALL BEARING DESIGN 159 

7. What should be the bearing surface on the V's of a planer table 
if the table and work weighs 45000 lbs., and the maximum pressure 
produced by cutting is 700 lbs., with a maximum rate of table travel 
of 2 ft. per second, using ring oil lubrication and allowing a factor of 
safety of 4? 

8. If 24 square inches is required for project area of a bearing, 
what will be its size according to formula (X. : D : : (i>/2V) : 1), if the 
number of revolutions per min. is 300. 

9. What should be the length of two bearings on a lathe if an \\" 
shaft is used, and the weight of cone pulley, shaft, and pressure of maxi- 
mum cut equals 1000 lbs., providing 3/5 of the pressure is on the front 
bearing, and the maximum number of revolutions per min. will be 1200? 

10. What must be the length of two bearings to support a 10 ton fly 
wheel on a 7" shaft running at 150 r.p.m., providing the wheel is 2 ft. 
from the center of one bearing and 3 ft. from the center of the other 
bearing, if force feed lubrication is to be used? 

Ball Bearing Design 

Ball Bearings can be divided into 3 groups; Radial, those 
that carry a radial load; Thrust, those that take a thrust 
or end load and the Combination Radial and Thrust, which 
take both a radial and thrust load. 

Ball bearings are especially adapted for high speed and 
light running machines. They are used in preference to 
sliding bearings due to the following reasons: less loss of 
power on account of the smaller coefficient of friction, less 
danger of bearing heating, and also because shorter bearing 
can be used. 

Rolling friction is always less than sliding friction. 

Friction of balls is independent of viscosity of the lubricant 
used. 

Ball bearings should always be lubricated, preferably with 
hard oil. Nothing but tool steel or alloy steel should be 
used for high grade, work although case hardened machine 
steel balls are sometimes used for large balls and light running 
conditions. 

The carrying capacity of a ball bearing is directly propor- 



i6o 



INDUSTRIAL MATHEMATICS 



tional to the number of balls and to the square of the ball 
diameter. 

The Permissable Load that annular ball bearing will carry 
can be determined approximately by the following formulas: 

P 
P = 0.65 X Y X £> 2 X N, N 



0.65 XFXD 2 



D = \~ 
Vo. 




Fig. I. Radial Type 



.65 X Y X N 

P = load capacity in pounds, 

D = diameter of balls, taking J" as a 

unit of diameter. Example 

(f" D = 3), 
N = number of balls, 

Y — constant which varies with condi- 

tions and type of bearings, 
also material and speed. 

Y = 10 for 500 r.p.m. 

Y = 7.5 for 1000 r.p.m. 

Y = 5 for 1500 r.p.m. 



EXERCISES 

1. What is the allowable load on a ball bearing having 10 — \" balls 
on a shaft running at 1500 r.p.m.? 

2. How many \" balls are necessary in a ball race to carry a load of 
1,000 lbs.? — for 2,000 lbs.? (Shaft running at 1,000 r.p.m.) 

3. What size balls must be used to carry a load of 300 lbs. providing 
10 balls are to be used, shaft running at 1,000 r.p.m.? 

4. What size balls must be used in bearings of an electric motor if 
total load on the two bearings is 700 lbs., 12 balls to be used in each 
bearing and motor is running at 1500 r.p.m.? 

5. What size balls are required in each bearing if the load consists 
of a 900 lb. fly wheel, and a 267 lb. shaft running at 1,000 r.p.m. providing 
the fly wheel is 5 ft. from one' bearing, and 2§ ft. from the other, and 
50 balls are to be used in each bearing? 

6. How many x§" balls can be placed around a shaft which is 2" in 
diameter? 



CENTER OF GRAVITY, ETC. 



161 



7. What size balls must be used if I want to put 11 balls around a 
shaft 2.550" in diameter? 

8. How many §" balls can be put in a bearing if the outer ball race 
is 1.456" in diameter? 

Centers of Gravity and Gyration, and Moment of Inertia, etc. 

Center of Gravity of a body is the point about which, if 
suspended, all the parts will be in equilibrium, that is, there 
will be no tendency to rotate. The center of gravity of a 
figure, symmetrical about 'a center line, lies on that line. 
To find the center of gravity of any figure, draw a line B-C 
(Fig. I), and perpendicular limited lines B-E and F—C. 
Divide B-C into any number of equal parts, and erect 
perpendiculars a, b, c, d, etc. at their middle points. 

The area of each section is the product of the length of 
middle line multiplied by width of section. 




Fig. I 



Fig. II 



The Moment of Area about the axis E-B is found by 
multiplying each area by the distance from the axis E-B to 
the middle line. 

The sum of the total moment of area divided by the total 
area is equal to K, or the distance from axis E-B, to the 
center of gravity of the area. 

Sum of moments of area = total area times K, or 



K = 



sum of moments of area 
total area 



1 62 



INDUSTRIAL MATHEMATICS 



By turning the figure 90 deg. and repeating the process, 
L-M (Fig. II) is found. The intersection of these lines is 
the center of gravity of the whole figure. The greater the 
number of divisions, the more accurate will be the result 
obtained. The center of gravity of an irregular shaped body 
which lies in a plane, is found by the same method. 

Sum of all products of each weight of 

section multiplied by distance 
total weight 

The center of gravity of a body which does not lie in a plane 

may be found by taking the 
moment about 3. planes which 
are at right angles to each 
other. 

The Center of Gravity of 
a Figure if composed of sev- 
eral rectangles, triangles, etc. 
can easily be found in the fol- 
lowing manner: 

If A, B, C = area of rect- 
angle; a, a', b, b', c, c' = dis- 
tance the center of gravity of each rectangle is from axis X 
and Y, which was conveniently selected; x, y = distance the 
center of gravity of whole figure is from axis Xand Y. Then 

(A Xa) + (B Xb) +(CXc) 



<— . 


-x 


....... 


C 






A 


opz: 






A 






"~? ' 


i|T 




: c» 

1 i i 






1 : 

II 


j 


a 


L-j 


--> 


n 






Fig. Ill 



X = 



A + B + C 
_ {A Xa') + ( BXb') +(C X c') 
y ~ A + B + C 

Then the point of intersection of the lines at = center of 
gravity. 

The center of gravity of any triangle is located at the 
intersection of lines drawn through | of the altitude, and 
parallel with the bases. 



CENTER OF GRAVITY, ETC. 



163 



The Moment of Inertia of a body (or section) with respect 
to an axis, is the sum of the products obtained by multiplying 
the weight (or area) of each elementary particle by the 
square of its distance from the axis. It is generally repre- 
sented by the symbol /. The value of it varies according 
to the position of axis. The moment of inertia is smallest 
when the axis passes through the center of gravity. 

The moment of inertia of any section about any axis is 
equal to its moment of inertia about a parallel axis through 
the center of gravity, plus the product of the area and square 
of the distance between the axis. 

If I' = Moment of inertia about an axis through center 
of gravity of section. 
/ = Moment of inertia about any parallel axis. 
A — area of cross-section. 
r = distance between axis of V and 7". 
Then / = /' + A X r 2 . 



"j-/f". 



The value of moment of inertia of various sections about 
an axis, passing through center of gravity of section, is given 
in table V, (page 114). 

Example: Find the moment 
of inertia of the section of a 
beam shown in Fig. IV about 
the horizontal axis E-F through 
the center of gravity. Divide 
the figure into 3 rectangles A, 
B, C. The center of gravity of 
each rectangle will lie on a line 
running through the middle height. 



-4-b' 

7 — 



Fig. 
Therefore 



-! * 

6 X 

">T! 



H^ 



IV 



3// 
4 • 



distance of A from G-H = ij -5- 2 = 
distance of B from G-H = ij + 3/2 = 3". 
distance of C from G-H = i| + 3 + f = si". 



= 2 9 /i6". 



164 INDUSTRIAL MATHEMATICS 

Area of A = ij X 6 = 9 sq. in. 

Area of 2J = 3 X i| = 4J S Q- m - 

Area of C = i| X 3 = 4i sq. in. 

Total area = 18 sq. in. 

sum of moments 

x = 

total area 

_ (9 X f ) + ( 4 j X 3) + (4§ X 5l) 
18 
Distance of center of gravity A from E-F (a') 

= 2 9/16" - f" = I 13/16". 
Distance of center of gravity B from E-F (b') 

= 3" - 2 9/16" = 7/16". 
Distance of center of gravity C from E-F {c') 

= 5 J -29/16" = 211/16". 

Moment of inertia of rectangle about its neutral axis (axis 
through center of gravity) is bd 3 /i2 (from table V, page 114). 

6 X (1J) 3 

Therefore moment of inertia of A = la = — — = 1.69, 

12 

t • • r D n. J J X (3 )3 
moment of inertia of B = lb = = 3.38, 

12 

r • • r n r 3 X (l| ) 3 

moment of inertia of C = Ic = = 0.84. 

12 

Moment of inertia about the new axis E-F is: 

I'a = la + 9 X (1 13/16) 2 = 1.69 + 29.51 = 31.20. 
I'b = lb + 4} X (7/16) 2 = 3-38 + 0.86 = 4.24. 
I'c = Ic + 4 | X (2 n/16) 2 = 0.84 + 30.55 = 31-39- 
Required total moment of inertia 

= I'a + I'b + I'c = 66.83 (Ans.). 

The Center of Gyration with respect to an axis, is a point 
at which, if the entire weight of a body is concentrated, its 
moment of inertia will remain unchanged; or in a revolving 



RADIUS OF GYRATION 



165 



body the point in which the whole weight of the body may 
be considered to be concentrated. 

Radius of Gyration is the distance from the axis to the 
center of gyration. 

/ = moment of inertia of a body. 
V = moment of inertia of a section. 
W = weight of the body. 
G = radius of gyration. 
A = area of section. 

7 = w X G\ 
r = A X G\ 

The value of radius of gyration for several sections can be 
found from table V (page 114). 



EXERCISES 

1. Find the distance of center of gravity of an angle section 6" X 4' 
X 1", from the back of each leg (Fig. I). 



v~ 


<— 










* 






\ 


<- 


-4 


<--?»- 


- > 


A 


Y 


£"- 


> 


...Y 






* 



Fig. I 



Fig. II 



2. Locate the center of gravity of the figure shown in Fig. II, taking 
the origin at the left hand corner of the bottom. 

3. A rectangular board 10" wide and 12" high has a circular hole cut 
out. Center of hole is 4" from the right edge and 4" from the bottom. 
Its diameter is 3". Locate the center of gravity of the remainder, 
taking origin at the lower left hand corner. 

4. Find the moment of inertia of the angle of problem No. 1, with 
respect to an axis parallel to its base through its center of gravity. 

5. Also of Fig. II. 



PART IV 

Graphical Charts 

In many instances records, statistics, data, etc., of in- 
dustrial organizations are kept by means of graphical charts, 
which provide a quick and accurate method of tabulating; 
same also being valuable to the busy executive, who can 
find at a glance, the information desired. 

Charts of this nature are also used frequently by statis- 
ticians, engineers, production managers, and experimental 
laboratories for keeping records, plotting values, etc. 

Graphical charts may be constructed in many forms, 
depending upon the nature of the work to be tabulated. 

The following are a few of the methods generally used and 
should prove of value to those who desire to become familiar 
with keeping records graphically. 

Fig. I shows a chart which is frequently used in keeping 
track of production. The chart is outlined in advance 
according to the production schedule desired; for instance, 
if the management decides that they desire to manufacture 
3000 machines between the months of January and October 
inclusive. The chart is divided into months as shown hori- 
zontally together with the number of parts desired per 
month. In this chart each month is divided into 3 divisions 
representing 100 parts each, although any other convenient 
scale may be used. On the left hand side and vertically are 
placed the name and number of the parts. As the parts are 
machined in the shop the production clerk blocks in the space 
opposite each part, thus the length of the blocked line after 
each part shows at a glance the total production up to date. 

For instance in the month of April 1200 gears No. 3876 

166 



GRAPHICAL CHARTS 



I6 7 



have been made which is just even with the advanced 
schedule, while only 1100 levers No. 4401 have been made 
showing a shortage of 100 thus it will be necessary to speed 
up production on this part. The chart shows that 2400 
pins No. 5628 have been made and as this is sufficient to last 
until August thus the machine or machines making these 
pins can be set to work on some other parts for a few months. 
Charts of this form are usually fastened to a board on the 
wall and are filled in by means of a color crayon or in some 
cases tacks are moved along a line opposite the parts as the 
production advances. The location of the tacks on the 
chart shows the maximum production up to date. 



NAME NO. 

Gear 3876 1 

Lever 4401 

Rod 7783 

Bushing 3Z74 

Pinion 9?35 

Pin sm 



£ o _o o 

O o uo 




Fig. I 



Fig. II is also an advance production chart although it 
can be used for many other purposes. 

Thus if a concern decides to make 1200 adding machines 
in one year a chart may be plotted as shown; lines repre- 
senting months are drawn vertically and lines representing 
quantities of 100 per division, more or less as desired, are 
drawn horizontally. A line connecting the lower left hand 
point with the upper right hand point will be the theoretical 
production line. As shown January production was 30 
machines, February 70 machines, thus the sum of the 
months production is always added to the total sum of the 
production up to the preceding month to obtain the location 



1 68 



INDUSTRIAL MATHEMATICS 



of the production point for the month in question. 70 
machines being made in February the production point will 
be 30 plus 70 equals 100 machines or total production up to 
the end of February. In March 75 machines were made, 
thus 100 plus 75 equals 175 or the location of the production 
point for March. By placing the production points in for 
the various months and connecting same a production curve 
will be generated, which will show the number of machines 
made each month, the total number of machines made up 
to any month and it will also show how close the actual 
production curve follows the theoretical one. 



1200 
1100 
1000 
900 
800 
700 
600 
500 
400 
300 
?00 
1 00 










































**?' 






































+* 




































*' 


.*»"' 


































.-' 


<•'' 




































X 








































































^y 


s' 


/ 
































*' 


s* 


































,' 


s' 




































y* 




































-'' 




































^' 


,^ 







































Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec 



Fig. II 



Fig. Ill shows a monthly chart which is laid out for a 
daily record of the number of cars assembled and shipped 
in a large automobile plant. The body of the chart is laid 
out the same as Fig. II after which the daily reports can be 
located on the chart and lines connecting these points will be 
a broken curve showing the daily fluctuation. 

To distinguish one curve from another a combination of 
broken lines, as shown, can be used or if desired inks of 
various colors can be used which will give a pleasing contrast. 



GRAPHICAL CHARTS 



I69 



Fig. IV is a chart somewhat different from the preceding 
ones and is very simple and easy to understand. It can 
be laid out vertically or horizontally as desired with the 
overall length representing the whole or 100%. Then, by 
dividing the overall length into lengths of such proportion 
as are desired and shading or coloring each proportion differ- 
ently a very effective chart is produced. 



100 

90 

80 

1. 70 

■§50 

3 40 

Z30 

20 

10 





-«— 7- S 


" r 3 s^ 7 


a 17 V a _ >t^ 7 Z\ s s_«vs^ -H 


LX.MV IAZ SZS_/^- 5 ^2 *-5 


WW^ "* S2^ 


JiZ 




— — Cory assembled 


Cars sh\ppecf 


Mil 



II 3 4 5 6 



9 10 II 12 13 15 16 17 18 
Date 

Fig. Ill 



20 22 23 24252627 29 3031 



This chart shows the proportion of the distribution of heat 
in a heating plant. 

Thus 16% of the heat produced by burning coal escapes 
through the chimney, 2% by radiation, 3% carbon in the 
ashes, 2% by moisture in the coal, 2§% unaccountable losses 
and the remainder or 74!% of the heat goes to heat the 
water in the boilers or called efficiency. 





Moisture in Coat 
Unaccountable Losses 



[ I Efficiency 

Fig. IV 



170 



INDUSTRIAL MATHEMATICS 



Fig. V is a rather peculiar form of chart which is easily 
constructed and shows the distribution of the units very 
effectively. The proportion of the chart that is of most 
importance is usually turned to face in one direction as 
shown (25% work) while the remainder of the proportions 
face in the opposite direction. 



100 

90 

80 

^•70 

S&0 

u 50 

«40 

30 

?0 

10 





25 PerCent Work 




25 Per Cent. Exhaust 



35 Per Cent Jacket Water 



\> "? PerCent Radiation and 
unaccountable Losses 

Fig. V 

This chart shows the distribution of heat in the average 
gasoline engine. Thus the total heat as developed by 
burning fuel is distributed as follows: 25% work, 13% fric- 
tion, 25% exhaust, 35% through the water which circulates 
around the cylinders, and 2% due to radiation and unac- 
countable losses. 

Fig. VI is a chart which was used in a press room although 
applicable in a variety of ways, so that the superintendent 
could tell the proportion of time the various presses were in 



NigW 



Noon 



11 > - Noon 
Machine 

No. \l I I 3 4 5 6 7 8 9 10 II 12 I I Z 4 5 6 7 8 9 10 II \Z I I I 4 5 6 7 8 9 10 II \l 



Night 




P-IZ 
P-14 



\Operatina 



\ldle 



■fi ' ! ng up \;:i'-\ Pepa/rs 
Fig. VI 



I Waiting ibr Stock 



GRAPHICAL CHARTS 171 

actual operation or whether they were idle, or inoperative 
due to setting up, repairs or waiting for stock. 

The so-called overtime or night portion of the chart was 
underlined heavily, as shown, so that any overtime would 
show up more distinctly as it was desired to keep the over- 
time work down to a minimum. 




Fig. VII 

No. Operation Time 

i. Place in chuck 10 

2. Start machine 5 

3. Face off ends 40 

4. Turn diameter 60 

5. Cut groove 38 

6. File off corners 10 

7. Stop machine 5 

8. Gauge piece 8 

9. Remove from chuck 4 

Total = 1.80 

Fig. VII is a circular form of chart which is used quite 
extensively for graphical representation of proportions. The 
circumference is usually subdivided into the proportions 
desired, after which lines are drawn from these points toward 
the center. These wedge shaped divisions afterward being 



172 



INDUSTRIAL MATHEMATICS 



numbered showing the unit of time or percent which they 
represent. Frequently the wedge shaped divisions are 
colored in various shades to bring out distinction. 

Fig. VIII is another form of circular chart although in 
reality it is form of chart similar to Fig. Ill developed 
around a common center. This chart was used by an 
employment manager and showed the total number of men 
hired per month. As shown in the chart, 50 men were hired 
during the month of January, 65 in February, 60 in March, 
etc. 



Jan 




Fig. IX is a chart plotted from a well-known 6 cylinder 



GRAPHICAL CHARTS 



173 



automobile gas engine having a 2>\" bore and a 5" stroke to 
show the horse power and torque in foot pounds at different 
speeds ranging from 200 to 2400 r.p.m., equivalent to a car 
velocity of 4.3 to 51.7 miles per hour respectively. 

As shown, the h.p. and torque were plotted at steps of an 
increase of 200 revolutions each and a curve drawn through 
these points gave the h.p. and torque curve from which the 
h.p. or torque can be obtained for all intermediate speeds. 



45 



40 



35 



30 



Z5 



140 
135 
130 

% 20 a) 120 

o" c-115 

r 15/5110 

.105 

10 3100 

95 
5 90 

85 
80 















1 — 
































































































a 


ir^ 
















































i 


irC} 


ue 


Curu 


f e- 






— 




































































































































































































_. 


































































x 


¥l 






— i 


**"■*■ 


^ 


^ ^ 






























/ 


s 














v 


N 


*>,. 
























1 


/ 




















V 


% 




















1 


/ 


























\ 


















/ 




























> 


^ 














































> 


\, 








/ 


* 




































V 


^ 






1 








































> 


k 














































>: 


) 































































































o o 
o o 



OOOOO OOOOOO 
OOOOO OOOOOO 



't LO ^S r- 00 cr> O 



8 



000000000 
000000000 



OJ l~0 rf Lf) vS r- OO c< o 



Revolutions per Minute 



oj m- 1x1 



(u W M M M M w 

Miles per Hour 
Fig. IX 



co (T- — 0-1 

O' Ki IX) 

"*- ■* ■* 



•^- id r— 



EXERCISES 
1. Draw a production chart similar to Fig. I for the months of 
January to August inclusive for 4 parts, a crank, gear, axle and con- 
necting rod, if the desired output is 500 parts of each per month. In 
the chart show a production of a full number of cranks for the month 
of May, 300 gears and 900 axles in excess for May and 250 behind 
schedule on connecting rods. 
13 



174 



INDUSTRIAL MATHEMATICS 



2. Lay out a chart similar to Fig. II showing a theoretical production 
curve for a period of 6 months, if it is desired to make 12,000 three inch 
shells. Also show the actual production curve from the following data: 
January 500, February 1600, March 1800, April 2500, May 2000 and 
June 3600. 

3. Make a monthly chart for an employment office similar to 
Fig. Ill for a period of one year, showing two curves, one for skilled 
and the other for unskilled labor, if the following number of each were 
hired per month. 



Month 


Skilled 


Unskilled 


Month 


Skilled 


Unskilled 


Jan 


28 
30 
16 
40 
70 
60 


22 
40 
28 
50 
38 
78 


July 

Aug 

Sept. , 

Oct 


IOO 
82 
68 
54 
50 
38 


120 


Feb 


98 
IIO 

80 


Mar 

Apr 


May 

June 


Nov 

Dec 


63 
71 



4. Lay out a diagram similar to Fig. IV showing the proportion of 
the number of various automobiles as stated below in the state of Michi- 
gan. 



Fords — 40 percent 
Buick — 7 percent 
Overlands — 6 percent 
Studebaker — 5 percent 



Reo — 4 percent 
Maxwell — 3 percent 
Cadillac — 3 percent 
All others — 32 percent 



5. Draw a diagram similar to Fig. V showing the proportion of the 
various substances as shown by the following analysis of the average 
run of Pennsylvania anthracite. Free carbon 82.5 percent, ash 9.0 
percent, volatile matter 5.0 percent and moisture 3.5 percent. 

6. Lay out a chart similar to Fig. VI showing the average time 
you spend per 24 hours at the following: sleep, meals, recreation 
work and idleness. 

7. Make a circular chart similar to Fig. VII showing the ratio of 
the proportions of the following analysis of a bearing alloy: copper 
74 percent, zinc 11 percent, lead 10 percent and tin 5 percent. 

8. Make a chart similar to Fig. VIII showing the average monthly 
temperature for a period of one year from the following data: 



Jan. -15 Feb. -8° March -2 ° April 12 May 40 June 72 ° 
July 86° Aug. 92 Sept. 8o° Oct. 76 Nov. 70 Dec. 35° 



STRENGTH OF MATERIALS 



175 



9. Plot a horse power and torque curve similar to Fig. IX from the 
following data obtained from a 6 cylinder gas engine in a dynamometer 
test. 



1 

R.P.M. 


H.P. 


Torque in 
Ft. Lbs. 


R.P.M. 


H.P. 


Torque in 
Ft. Lbs. fc 


200 


2.5 


95 


1400 


36 


131 


400 


7 


115 


1600 


38 


126 


600 


15 


126 


1800 


42 


120 


800 


21 


135 


2000 


44 


110 


1000 


26 


142 


2200 


43 


94 


I200 


31 


136 









10. Construct a temperature curve from the following figures which 
were the temperatures of a piece of tool steel placed in a furnace brought 
to a temperature of 1600 and taken out and allowed to cool. Tempera- 
tures were taken after the first minute and every minute thereafter for 
20 minutes. 400, 550, 730, 910, 1075, 1240, 1350, 1400, 1410, 1475, 
1580, 1600, 1510, 1360, 1280, 1260, 1120, 975, 700 and 450. 



Strength of Materials 

A Stress is a force which acts in the interior of a body, 
and resists the external forces which tend to change its shape. 

The Unit Stress is the stress per unit of area, usually per 
square inch. It is equal to the total stress divided by the 
area of the cross section in square inches. 



Formula: p = — or P = A X p. 

A 



(I). 



Example 1 : Find the unit stress in a bar of iron 2" square 
under a load of 3200 lbs. 
Formula (I): 

P 



3200 3200 



2X2 



= 800 lbs. per sq. in. (Ans.). 



Example 2 : Find the total stress in pounds produced on a 
6" X 6" sq. oak timber, which is under a unit stress of 250 
lbs. per square inch. 



176 INDUSTRIAL MATHEMATICS 

Formula (I): 
P = A X p = (6 X 6) X 250 = 36 X 250 = 9000 lbs. (Ans.). 

A Strain is the deformation or alternative produced on a 
body by external forces. When external forces act on a body 
they produce stresses within the body. 

Stresses are of five kinds: Tensile, Compressive, Trans- 
verse or Bending, Shearing and Torsional. In most cases a 
Combination of these stresses is produced. Any stress (how- 
ever small) in a body, produces a deformation or alternation 
in the shape of the body. 

If the stress is not too large, the body will return to its 
original shape and size when the external force is removed. 
This property of a body which enables it to return to its 
original, size and shape, is called its Elasticity. 

If the force be great enough the body will not return to its 
original size or shape when released, but assumes a new 
shape, which is called a set, and we say, "its Elastic Limit 
has been exceeded." 

Up to the elastic limit the deformation in a body is directly 
proportional to the load, and the elastic limit may be defined 
as that point beyond which the deformation ceases to be 
proportional to the load, or the point at which the rate of 
stretch begins to increase. 

The Modulus (or coefficient) of Elasticity is the relation 
between the amount of extension or compression of a body, 
and the force which reduces that extension or compression. 
It may be defined as the load per unit of section, divided by 
the extension per unit of length, and is the quotient obtained 
by dividing the stress per square inch by the elongation in 
one inch caused by this stress. 

PL PL 

Formula: E = or e = . (II). 

Ae AE 



STRENGTH OF MATERIALS 177 

Example 3: Find the modulus of elasticity of a 2" sq. 
steel bar 10 ft. long which elongates 0.080" under a load of 
80,000 lbs. 

PL 

Formula (II): E = . 

Ae 

P = 80,000 lbs. 

L = 10 X 12 = 120 inches. 

A = 2 X 2 = 4 sq. in. 

e = 0.080". 

80,000 X 120 

E = = 30,000,000 (Ans.). 

4 X 0.080 

Example 4: How much will a bar of cast iron 3" sq. and 
20" long be compressed under a load of 80,000 lbs. 

PL 

Formula (II): e = . 

V AE 

P = 80,000 lbs. 

L = 20 in. 

A = 3 X 3 = 9 sq. in. 

E = 15,000,000. (Taken from table III, page 184.) 

80,000 X 20 

e = = 0.0118" (Ans.). 

9 X 15,000,000 

Example 5: Find the elongation of a 1" sq. steel bar 6' o" 
long under a load of 100,000 lbs. 

PL 

Formula (II): e = . 

AE 

P = 30,000 lbs. per sq. in. 
L = 6 X 12 = >]2". 
A = 1 sq. in. 

E = 30,000,000. (Taken from table III, page 184.) 
30,000 X 72 



1 X 30,000,000 X 1 



= 0.072" (Ans.). 



178 INDUSTRIAL MATHEMATICS 

The Breaking or Ultimate Stress of any material is the 
stress or force which will rupture it. 

The ultimate stress in tension, compression and shear for 
various materials is given in table II, page 183. 

Formula (III): P = A X S. 

Example 6: What will be the force in pounds necessary 
to rupture a steel test bar in tension which is 0.505" in 
diameter? 

Formula (III): P = A X S. 
A = 0.200 sq. in. 

5 = 100,000 lbs. per sq. in. (Taken from table II, page 183.) 
P — 0.200 X 100,000 = 20,000 lbs. (Ans.). 

Example 7: What force in pounds will be necessary to 
punch ai" hole in a steel plate \" thick? 

Formula (III): P = A X S. 

A = 1 X 3-H l6 X i = 0.7854 sq. in. 

5 = 70,000 lbs. (Taken from table II, page 183.) 

P = 0.7854 X 70,000 = 54,978 lbs. (Ans.). 

Example 8: What will be the weight in tons necessary to 
crush a brick pier 10" square. 

Formula (III): P = A X S. 
A — 10 X 10 = 100 sq. in. 

S = 2500 lbs. (Taken from table II, page 183.) 
P = 100 X 2500 = 250,000 -7- 2000 = 125 tons (Ans.). 

The Working Stress is the stress or load which a body will 
easily sustain or it is equal to the ultimate stress divided by 
the factor of safety. 

Formula (IV) : Working Stress = - . 

The Factor of Safety is the number by which the breaking 
stress must be divided to obtain the working stress. The 



STRENGTH OF MATERIALS 179 

factor of safety in ordinary practice varies from 5 to 30. 
This depends somewhat upon the formula used,, workman- 
ship, character of the load etc. (see table I, page 183). 

Example 9: Find the working stress in a beam which has 
an ultimate strength of 150,000 lbs. per sq. in., if the factor 
of safety used is 15. 

Formula (IV) : Working Stress = - . 

5 = 150,000 lbs. 

/ = 15. 

150,000 

Working Stress = = 10,000 lbs. per sq. in. (Ans.). 

15 

For tension, compression (where length does not exceed 
10 times its least diameter) and shear 

AS Pf 

P = — or A = — . Formula (V). 
/ 5 

Note. — When the length of a bar under compression is 
longer than 10 times its least diameter, it must be treated 
as a column and formula No. (VI) must be used. 

Example 10: Find the total stress in pounds imposed 
upon a cast iron bar 3" square, in tension, if the load is 
steady. 

AS 
Formula (V) : P = — . 
/ 

A = 3 X 3 = 9 sq. in. 

>S = 20,000 lbs. per sq. in. = ultimate strength of cast iron 

under tension. (Taken from table II, page 183.) 
/ = 6 for cast iron under a steady load. (Taken from 

table I, page 183.) 

9 X 20,000 
P = — = 30,000 lbs. (Ans.). 



l80 INDUSTRIAL MATHEMATICS 

Example n: What will be the safe load to apply under 
tension on a wrought iron bar 2" sq. under a varying stress? 

AS 

Formula (V): P = . 

/ 
A = 2 X 2 = 4sq. in. 

S = 50,poo lbs. (Taken from table II, page 183.) 
/ = 6. (Taken from table I, page 183.) 

4 X 50,000 

p = = 33,333! lbs. (Ans.). 

o 

Example 12: Find the size of square cast iron bar neces- 
sary to use to resist the compressive stress under a load of 
7 \ tons? 

Pf 
Formula (V) : A = — . 

P = 7.5 X 2000 = 15,000 lbs. 

/ = 6. (Taken from table I, page 183.) 

S = 90,000 lbs. (Taken from table II, page 183.) 

A = — '- = 1 sq. in. Therefore each side 

90,000 

= V7 77 = j" (Ans.). 

Example 13: What should be the diameter of a wrought 
iron bar to suspend a weight of 50,000 lbs.? 

Formula (V): A = — . 
S 

P = 50,000 lbs. 

/ = 4. (Taken from table I, page 183.) 

£ = 50,000 lbs. per sq. in. (Taken from table II, page 183.) 

50,000 X 4 

A — = 4 sq. in. or 2.257" diameter (Ans.). 

50,000 

Example 14: What should be the diameter of a steel 
journal in a machine to resist a shearing load of 10,000 lbs. 
safely? 



STRENGTH OF MATERIALS l8l 

Pf 
Formula (V) : A = — . 

P = 10,000 lbs. 

/ = 15. (Taken from table I, page 183.) 

S = 70,000 lbs. (Taken from table II, page 183.) 

10,000 X 15 . . 

A = = 2.142 sq. in. or approximately 

70,000 

1 21/32" diameter (Ans.). 

To find the breaking strength of a column, i.e., a vertical 
member whose height is greater than 10 times its least diam- 
eter the following formula is used. 
SA 



Formula (VI): P = 



U 
* G 2 



Example (15): Find the breaking load of a round wrought 
iron bar 2" in diameter and 3 ft. long, under compression, 
if both ends are rounded. 

SA 

Formula (VI) : P = . 

L 2 

* G 2 

5 = 50,000 lbs. per sq. in. (Taken from table II, page 183.) 

A = 2 X 2 X 0.7854 sq. in. = 3.1416 sq. in. 

4 

q = . (Taken from table VI, page 186.) 

36,000 

d 2 2 2 4 
G 2 = — or — or — . (Taken from table V, page 185.) 
16 16 16 

L = 3 X 12 = 36 inches. 

50,000 X 3-H l6 
P = — = 99,670.05 lbs. (Ans.). 

36,000 4 
16 



1 82 INDUSTRIAL MATHEMATICS 

In finding the breaking strength of a beam, i.e., a bar in a 
horizontal position the bending moments must be taken 
into consideration, and the following formula may be used: 

Formula (VII): M = S X R. 

In practice to find the safe working strength of a beam 
the proper factor of safety must be used which transforms 

Formula (VII) into M = - X R. Formula (VIII). 

Example 16: Find the maximum load a 3" square wooden 
cantilever 2 ft. long will support at its free end. 

S X R 
Bending moment = — = W X L. 

(Taken from table IV, page 184.) 

S X R 

.'. W = Formula (IX). 

fXL 

S = 9000 lbs. (Taken from table II, page 183.) 

d z 3 3 27 
jR = — = — = — or 42. (Taken from table VI, page 186.) 
6 6 6 

/ = 8. (Taken from table I, page 183.) 

I = 2 X 12 = 24 inches. 

9000 X 4J 40,500 

W = = = 210.9 lbs. (Ans.). 

8 X 24 192 

Example 17: Find the maximum bending moment pro- 
duced on a simple beam 10 ft. long loaded at the center with a 
800 lb. weight. 

W X L 
Formula: . (Taken from table IV, page 184.) 

W = 800 lbs. 

L = 10 X 12 = 120 inches. 
800X120 



Maximum bending moment = 



4 

= 24,000 inch lbs. (Ans.). 



STRENGTH OF MATERIALS 



I8 3 



The Deflection of a Beam (J) at the critical point or 
maximum deflection can be found from table IV according 
to the kind of beam and manner of loading. 

Example 18: What is the deflection of a rectangular steel 
beam fixed at both ends, 5 ft. long, 2" wide, 3" deep, sup- 
porting a 20,000 pound load in the center? 

Formula (X) : 6" = — , (Taken from 



I92£ X / 
page 184.) 
W = 20,000 lbs. 

L = 5 X 12 - 60". 

E = 30,000,000. (Taken from table III, page 1 



table IV, 



I = 



bd 3 2 X 27 



= 4J. (Taken from table V, page 185.) 



5 = 



20,000 X 60 3 



192 X 30,000,000 X 4 J 



0.0833" (Ans.). 



TABLE I 
Factor of Safety "/' 



Material 


Steady 
Stress 


Varying 
Stress 


Shocks 
(Machines) 


Cast iron 


6 
4 
5 
8 
15 


IS 
6 

7 
10 

25 


20 


Wrought iron 


10 


Steel 


15 
15 
30 


Wood 


Brick and stone . . 







TABLE II 
Ultimate Strengths "S" 



Material 



Tension 



Compression 


Shear 


90000 

5000O 

150000 

8000 


20000 
47000 
70000 
600 to 


6000 


3000 


2500 





Bending 



Cast iron .... 
Wrought iron. 

Steel 

Wood 



Stone . 
Brick. 



20000 

50000 

100000 

10000 



36000 
50000 

120000 

9000 

2000 



1 84 



INDUSTRIAL MATHEMATICS 



TABLE III 



Material 



Coefficient of 
Elasticity (E) 



Elastic 
Limit for 
Tension 



Cast iron 
Wrought iron 

Steel 

Wood 



15000000 

25000000 

30000000 

1500000 



6000 
25000 
50000 

3000 



TABLE IV 



Kind of Beam and Manner of Loading 



Bend- 
ing 
Mo- 
ment 
(M) 



Deflection 
(5) 



9 



000000, 



_Q 



QOOQQO 



D. 



^ OQQOOC W 
fea 



Cantilever, load at 
end 

Cantilever, uni- 
formly loaded .... 

Simple beam, load 
at center 

Simple beam, loaded 
uniformly 

Beam fixed at both 
ends, load at 
center 

Beam fixed at both 
ends, uniformly 
loaded ' 



WXL 


WXL 3 
ZE X 1 


WXL 


WXL 5 


2 


BE X I 


WXL 


W XL 



4 
WXL 



48£ X 1 
5^ XL 3 



8 



WXL 



384S X 1 



WXL 3 



8 



WXL 
12 



192E X 1 



WXL* 
3&4E X 1 



STRENGTH OF MATERIALS 
TABLE V 



185 



Name and Shape 


(/) 


(R) 


(G 2 ) 


Solid 
Circle 


e 




7T</ 4 
64 

64 


ird 3 
32 

7r(d 4 - di*) 
32d 


d2 
16 


Hollow 
Circle 


f(©) 


d 2 + <fl 2 
16 


Solid 
Square 




-d-> 




rf 4 
12 

d* - di* 

12 

&d 3 
12 

M 3 - fclrfl 3 
12 

bd 3 

36 


d 3 
6 

d 4 - di* 
6d 

bd* 
6 

bd 3 - bidi 3 
24 


d? 
12 


Hollow 
Square 


/<* 2 + </l 2 

V I2 - 

&2 
12 


Solid 
Rectangle 


r 

d 

i_ 

T" 

d 
i. 


td 


Hollow 
Rectangle 




d i 


/ &d 3 - &idx 3 
\ 12 (bd - bidi) 


Solid 
Triangle 




A 


i 


18 




P»b-*i 




Solid 
Ellipse 


© 


■xbd 3 

64 


irbd 2 
32 


ft 2 

16 


Hollow 
Ellipse 


If], 


tt(^ 3 - Ml 3 ) 
64 


7r(&<i 3 - &i<ii 3 ) 
32d 


6 3 i - &i 3 rfi 
16 (6d - bidi) 


I 




&d 3 - bidi 3 
12 


bd 3 - bidi 3 
6d 


b 3 d - bi 3 di 


Beam 


if |«- — Y,.—A 


12 (bd — bidi) 


Even 
Cross 


§> 


Area X </ 2 
19 


Area X d 
9-5 


d* 

22.5 


Even 
Angle 


EP 


Area X d? 
10.2 


Area X d 

7.2 


rf2 
25 



1 86 



INDUSTRIAL MATHEMATICS 



TABLE VI 



Material 


Both Ends 
Flat or Fixed 


One End 
Round 


Both Ends 
Round 


Cast iron 


I 
5000 

I 


1.78 
5000 
I.78 

36000 
I.78 

25000 
1.78 

3000 


_4_ 




Sooo 
4 


Steel 


36000 

I 


36000 
4 


Wood 


25000 
I 


25000 

4 




3000 


3000 



If the weight of the beam is large in comparison to the 
load on the beam, then it must be considered as a load uni- 
formly distributed over the whole length of the beam (pro- 
vided the beam is of uniform cross-section throughout). 

A Simple Beam is one that is merely supported at the ends. 

In the expression for (R) (table V) "d" is always under- 
stood to be the vertical side or depth. 

The moment of inertia (/) (table V) is taken about a plane 
perpendicular to "d" and lying in the same plane. 

EXERCISES 

1. A square cast iron pillar 18" long is required to sustain a steady 
load of 75000 lbs. What must be the length of a side? 

2. What would be the elongation of a round wrought iron bar 24" 
long and \\" in diameter under a tensile stress of 15 tons? 

3. What is the breaking load of a cast iron simple beam, uniformly 
loaded, which is 20 ft. long between supports, and of hollow rectangular 
cross section 6" X 8" outside and 4" X 6" inside, neglecting its own 
weight? 

4. What is the deflection of a solid wrought iron beam 6" wide and 
2!" deep, fixed at both ends, 7 ft. between supports, with a load of 21000 
lbs. in the center? 

5. What is the depth "d" of a steel cantilever 36" in length, that 
will sustain a weight of 4000 lbs. at the end, if beam is rectangular and 
2§" wide, neglecting its own weight? 

6. What is the breaking load on an elliptical wooden column, 
having rounded ends, the diameters of the cross section being 12" and 
8" and the column 18 ft. long? 



SPRINGS 



187 



7. A wrought iron bar is to support in tension a load of 20 tons, 
variable. Find the diameter of the bar required. 

8. What would be the elongation of the bar in problem No. 7 if the 
length was 5 ft.? 

9. A square bar firmly held at one end, is supporting a load of 
3000 lbs. at the outer free end. If the load is steady, the bar made of 
structural steel and is 2\ ft. long, what size should it be? 

10. If an elongation of 0.015" is produced in a steel bar 10" long, 
2" square by a tensile load of 90 tons, what is the modulus (or coefficient) 
of elasticity? 

Springs 

Springs are divided into three general classes: elliptical, 
helical and spiral. 

A Spiral Spring is one which is wound around a fixed 
point or axle, the diameter of each coil gradually increasing. 
(Fig. I.) 

A Helical Spring is one which is wound around an arbor, 
and at the same time advancing like the thread of a screw. 




Fig. 



Spiral Spring 




Fig. II 



(Fig. II.) A good proportion for this kind of spring is to 
make the mean diameter from 5 to 10 times the diameter of 
wire used. 




Half Elliptical 




Full Elliptical 



Fig. II] 



1 88 INDUSTRIAL MATHEMATICS 

An Elliptical or Laminated Spring is made up of flat bars, 
plates or leaves of uniformly varying length, placed one 
upon the other and held together by bolts and clips. (Fig. 
III.) 

Abbreviations Used in Formulas 
W = safe load in lbs. per sq. in., or maximum carrying 

capacity. 
W = load which causes the deflection 5' (where W is less 
than W) in lbs. per sq. in. 
5 = total deflection in inches. , 

b' = total deflection in inches under the load W. 
f = safe strength in lbs. per sq. in. (80,000 lbs. per sq. in. 

for ordinary spring steel). 
b = width of spring in inches. 
t = thickness of spring in inches. 
d = diameter of spring wire in inches. 
D = mean diameter of helical spring in inches. • 
H = free height of helical spring in inches (height without 

load). 
h = solid height of helical spring in inches (height when 

completely compressed). 
L = effective span = 2V. 
I = length of spring in inches. 
n = no. of leaves in half elliptical spring, or \ of total no. 

of leaves in full elliptical spring. 
R = length of load arm in inches. 
E = modulus of elasticity for tension and- compression 

= (30,000,000 for steel). 
G = modulus of elasticity for torsion = (12,600,000 for 

steel). * 
r = ratio of no. of full length leaves to total no. of leaves. 
Spiral Springs: 

fd s 20WIR 2 2flR 

\<dR Ed 4 Ed 

for round spring stock. 



SPRINGS 189 

fbt 2 12 WIR 2 2JIR 

62? EM* Et 

for rectangular spring stock. 

Example: Find the safe load for spiral spring 2" wide 
\" thick, 6 / -o' / long, when the load arm is 10". Also find 
the deflection 8 under the load. 

fbt 2 80,000 X 2 X (i) 2 

W = — = — -— = 166.7 lbs. (Ans. . 

6 X R 6 X 10 

2flR 2 X 80,000 X 6 X 12 X 10 

8 = J — = : = 15.36" (Ans.). 

Et 30,000,000 X \ 

Helical Springs: (a) round bar stock. 

G8d 5 wfd* 

W = = — , 

8hD* 8D 

irfhD 2 s , SW'hD> 

d = H - h = , 8 = „„ , 

Gd 2 Gd 5 



H = h+ F = h + 

G 

_ irhD _ H 



irfh/D^' 



;j 



d fD D 

h 0.32 — 

Gd 6 d 

(b) rectangular bar stock. 

fbt V^ + fe2 G8bt(t 2 + b 2 ) 

W = = , 

3 D 9 4hD* 

irfhD 2 H 

~ Gt ^f+T 2 ~ Gt V; 2 + b 2 ' 

1 + tfD 2 

9.dhD*W f 

8' = 



Gbt\t 2 + b 2 ) 
irhfD 2 



H = h + 

Gt \t 2 + b 2 



14 



190 INDUSTRIAL MATHEMATICS 

wDh H 



t fD . ■ , t 



G V/ 2 + b 2 D 

Example: Find the diameter and length of wire for a 

round helical spring, mean diameter i", solid height i", 

for a load of iooo lbs. 

tt/^3 *fcwD 3 / 8XioooXi „" . 

W = , d = \/ — ■ — = \ ■ — =0.179" (Ans.). 

SD \ tt/ \3.14X80000 

tt/lD 3-14 X 1 X 1 „. a 

/ = — — = = 1749 (Ans.). 

d 0.179 

Elliptical Springs: 

2 fnbt 2 
W = -- . 

3 L 

ifL 2 

d = for full elliptic. 

2 Et 

1 fL 2 

for full elliptic with more than one full leaf. 



2 + r Et 



* fL 2 

5 = for half elliptic. 

4 Et 

1 fL 2 
8 = for half elliptic with more than one full 

2(2 + r) Et 

leaf. 

Example: Find the safe carrying load and maximum deflec- 
tion of a semi-elliptical spring 2" wide, 0.266" thick, 9 leaves 
and 42" span, taking the value of / = 40,000 lbs. per sq. in. 

2 fnbt 2 2 40,000 X 9 X 2 X (0.266) 2 

W = — = - X 

3^3 42 

= 808 lbs. (Ans.). 

1 fL 2 1 (42) 2 

5 =--— = - X 40,000 X ; 

4 Et 4 30,000,000 X 0.266 

= 2.409" (Ans.). 




Vertical Milling Machine 









if 



<D Z y- I- 

<< | 2 

(-1 o ° 

£ 



r 
o 

H 

3* 
3 3 

m 
< 



QD 

X 



■7 



PIPES AND CYLINDERS 



191 



EXERCISES 

1. What is the deflection of a spiral spring 1" wide, f " thick and 
2'-o" long, under the maximum safe load which has a lever arm of 3"? 

2. What is the maximum carrying capacity of a \" round steel 
helical spring with a mean coil diameter of 1"? 

3. If the solid height of the above spring is 2|", what is the free 
height? 

4. Also find the deflection of the above spring under a load of 250 lbs. 

5. Find the diameter of wire required for a round helical spring 
under the following conditions: 

Maximum load 1 100 lbs. 

Mean diameter of coil i|" 

6. Find the total deflection of the above spring under a load of 600 
lbs. when the solid height is 4". 

7. A helical spring of \" steel wire is to have a compression of i\" 
under a load of 2000 lbs. What is the outside diameter and solid height 
of the coil? 

8. Also find the total length of the above spring. 

9. A semi-elliptical spring consists of 7 plates of f" thickness. 



the value of / = 80,000 per sq. in. E = 25,000,000. 

10. A semi-elliptical spring has 11 leaves of 9/32" thickness, 2" wide, 
with a span of 52". Find the safe load using/ = 40,000 lbs. per sq. in. 
E = 30,000,000. 

Pipes and Cylinders 

When a hemispherical vessel, suspended 
by a string as in Fig. I under internal 
steam pressure, will move neither to the 
right nor to the left, this proves that the 
total pressure on the curved surface in di- 
rection F is equal to that upon the flat sur- 
face in direction E. 

The flat surface is called the Projected 
Area of the curved surface. 




E-%- 



~F 



Fig. I 



192 INDUSTRIAL MATHEMATICS 

Abbreviations Used in Formulas for Cylinders 

d = inside diameter of cylinder in inches. 

/ = thickness of cylinder wall in inches. 

p = pressure in lbs. per sq. in. 

s = working stress of cylinder walls in lbs. per sq. in. 

/ = length of cylinder in inches. 

P — total pressure in one direction in lbs. per sq. in. 

Total pressure (P) is equal to (pdl). 

The resistance of the boiler to this pressure on each side 

A and B (Fig. II) is equal to (stl). 

Therefore: pld = 2stl. 

pd = 2St. 

2st pd 

Formula I : p = — or / = — . 
d 2S 

The pressure has also a tendency to rupture the receptacle 
circumferentially by pulling it apart lengthwise. (See 
Fig. III.) 




m 



Fig. II Fig. Ill 

The area of cylinder head against which the pressure acts 
is equal to jTd 2 and the total load exerted on the shell 
circumferentially is equal to the total area in sq. in. of one 
end, times pressure per sq. in. 

The area of metal to resist this pressure is irdt. The total 
resistance is wdts. Therefore iird 2 p = irdts. 

4ts pd 

Formula 2 : p = — or/= — . 
d 4s 



PIPES AND CYLINDERS 193 

The thickness of cylinder obtained by the formula (2) is 
just half of that of formula (1). Thus, the cylinder generally 
fails along the longitudinal section. 

Example: Find the thickness of a cast iron cylinder 4" in 

diameter to resist a 400 lb. pressure, taking ultimate tensile 

strength of C.I. as 20,000 lbs. per sq. in., using a factor of 

safety of 10. 

pd 400 X 4 
t =*-= Jt *- = 0.4" (Ans.). 

2S 2 X 20,000 
10 

Where boiler flues, tubes, etc. are under external pressure, 
this pressure has a tendency to change the shape of the 
round tubes into an elliptical cross-section. This distorsion 
when once begun, increases rapidly and failure occurs by the 
collapsing of the tube. 

Abbreviations Used in Formulas for Pipes and Tubes 

p = collapsing pressure in lbs. per sq. in. 
t = thickness of tubes in inches. 
d = outside diameter of tube in inches. 
/ = length of pipe in inches. 

( t\ 

Formula 3: p = I 87000- 1 — 1400. (For lap welded steel 
tube when p is greater than 600 lbs. per sq. in.) 

( A 3 

Formula 4: p = 50000000 I - I . (For lap welded steel 
tube when p is less than 600 lbs. per sq. in.) 

EXERCISES 

1. Find the thickness of steam cylinders 12" in diameter to resist 
the steam pressure of 120 lbs. per sq. in., using working strength of 
2000 lbs. per sq. in. 

2. Find the maximum steam pressure allowable for a cast iron 
cylinder 6" in diameter ^" thick, taking working strength as 2000 lbs. 
per sq. in. 



194 



INDUSTRIAL MATHEMATICS 



3. Find the thickness of a cast iron water pipe 6" in diameter, under 
a pressure of 300 lbs. per sq. in., taking working strength as 1500 lbs- 
per sq. in. 

4. Find the thickness of a gas engine cylinder 3" in diameter, assum- 
ing the maximum explosion pressure to be 400 lbs. per sq. in. (Use the 
formula No. 2 — / = 2000 lbs. per sq. in.) 

5. What is the collapsing pressure of a lap welded steel tube t§" 
thick and 2" in diameter? 

Riveted Joints 

Riveted Joints may be classified according to the method 
of connecting the plate and the number of rows of rivets used. 

Lap Joints are joints where main plates overlap each 
other, as in Fig. I (a and b). 

Butt Joints are joints where edges of main plate butt 
against each other and the connection is made through cover 
plates, as in Fig. II. 





(a) (6) 


Butt joint 


Single riveted Double riveted lap 


Fig. II 


lap joint joint 




Fig. I 





Several methods of failure 



C: 



) ~@~ 



(!) 
Shearing 



(2) 
Tearing 




(4) 
Bending 



(5) 

Bearing 



RIVETED JOINTS I95 



(2) Failure due to tensile stress of plate. 

(3) Failure due to shearing stress of plate. 

(4) Failure due to bending stress in the plate. 

(5) Failure due to bearing or compressive stress in the rivet. 

Abbreviations Used in Formulas 

ft = tensile strength of plate in lbs. per sq. in. 
f s = shearing strength of rivets in lbs. per sq. in. 
f c = compressive strength of rivets in lbs. per sq. in. 
t = thickness of plate in inches. 
d = diameter of rivet in inches. 

P = pitch of rivets in inches. (Distance from center of one 
rivet to another.) 
Formula for single riveted lap joint (Fig. l-a): 

ird 2 
Resistance to shearing one rivet = — f 8 . 

4 
Resistance to tearing between rivets = (P-d) t.ft- 
Resistance to compressing rivet or plate = d.t.f c . 



ird 2 

— /. = d.t.u 

4 


or 


fct 

d = 1.27 — . 

fs 


— ./, = (P-d) t.ft- 

4 


or 


Trd 2 f 8 

P = — - + d. 
4 tf t 



Formula for double riveted lap joints (Fig. l-b): 

2ird 2 

Resistance to shearing 2 rivets = f„. 

4 
Resistance to tearing between 2 rivets = (P-d) t.ft. 
Resistance to compressing in front of 2 rivets = 2d.t.f c 

2ird 2 f c t 

f 8 = 2d.t.f c ., d = 1.27 — , 

4 Js 

2wd 2 ird 2 f 8 

/. = (P-d) t.f t ., p= — J - + d. 

4 2 tft 



I96 INDUSTRIAL MATHEMATICS 

The efficiency of a rivet joint is equal to the ratio of the 
section of plate left between the rivets to the sections of 
solid plate; or the ratio of the clear distance between 2 
adjacent rivet holes to the pitch. 

Material Tensile Strength in Lbs. {ft) Compressive Strength in Lbs. (J c ) 

Wrought iron 50,000 80,000 

Steel 56,000 90,000 

Shearing Strength in Lbs. (/*) 

Wrought iron 40,000 

Steel 45,000 

EXERCISES 

1. Calculate the diameter and pitch of steel rivets, single riveted 
lap joint, for ye" steel plates. 

2. Find the diameter of rivets if the above plates are wrought iron. 

3. Find the diameter and pitch of steel rivet for \" wrought iron plate, 
double riveted lap joint. 

4. Find the efficiency of joints of the above problem. 

5. Determine the required number of steel rivets for a joint to carry 
20,000 lbs., using |" steel plate. 

6. If 2 plates 4" wide and §" thick are connected by two \" rivets, 
what load will the joint safely carry? 

Logarithms 

Logarithms are used to facilitate mathematical calculations 
involving multiplication, division, involution and evolution, 
which would otherwise require considerable labor and con- 
sequent risk of error. 

When a number (N) is equal to the ( X) power of a number 
(a), or N = a x , X is called the logarithm of the number (iV), 
and (a) is called its Base. It is expressed in the following 
way: log a N = X. 

Rule 1. — The logarithm of the product of two or more 
numbers is equal to the sum of the logarithms of several 
factors. Example: log a (M X N) = log a M + log a N. 

Rule 2. — The logarithm of the quotient of two numbers is 
equal to the remainder found by subtracting the logarithm 



LOGARITHMS 197 

of the divisor from the logarithm of the dividend. Example: 

log a M/N = log a M ~ loga N. 

Rule 3. — The logarithm of the power of a number is the 
product of the logarithm of the number multiplied by the 
exponent or index of the power. Example: log a N k = K 
log a N. 

Rule 4. — The logarithm of the root of a number is the 
quotient found by dividing the logarithm of the number by 
the index of the root. Example: log a ViV = i/k logo N. 

A logarithm consists of two parts, the whole number called 
the Characteristic and the decimal the Mantissa. Example: 
Log 1.2549. The figure (1) is the characteristic and the 
figure (2549) is the mantissa. 

When the base (a) of logarithms is 10, logarithms are called 
Common Logarithms. 

When the base (a) of logarithms is (e) or 2.7 1828 19 + , 
logarithms are called Natural, Hyperbolic or Napierian 
Logarithms. 

The common logarithm is used in ordinary calculations, 
and the hyperbolic logarithm is used in higher mathematics. 

In this book we treat only with common logarithms, as 

j . 3010 =2 or log 2 = 0.3010, 

I0 i.30io _ 20 or i g 20 = 1. 3010, 

IO 2.3010 = 200 or l Q g 200 _ 2> 30I0. 

The above figures show that the logarithms of 2, 20 and 
200 have the same mantissa, the only difference being the 
value of characteristics. The mantissa of any number from 
I to 1000 can be found from the table below. Therefore we 
must prefix a suitable characteristic to it afterward 

since io° = 1 therefore log 1 =0, 

io 1 = 10 log 10 = 1, 

io 2 = 100 log 100 = 2, 

io -1 = 1/10 = 0.1 log 1/10 = — 1, 

io~ 2 = 1/100 = 0.01 log 1/100 = — 2. 



I98 INDUSTRIAL MATHEMATICS 

It is evident that the common logarithm of any number 
between 

1 and 10 will be o plus a decimal, 
10 and 100 will be 1 plus a decimal, 
100 and 1000 will be 2 plus a decimal, 
1 and 0.1 will be — 1 plus a decimal, 
0.1 and 0.01 will be — 2 plus a decimal. 

The above figures show that the characteristics of all 
numbers less than 1 have a negative value, and the charac- 
teristics of all the numbers greater than 1 have a positive 
value. The mantissa is always made positive. 

When the characteristic has a negative value, it is custom- 
ary to write the minus sign over the top, or to add 10 to it 
and to indicate the subtraction of 10 from the resulting 
logarithm. Thus log 0.2 = 1.3010 or 9.3010 — 10. 

Examplei: Find the logarithms of 2056, 20.56 and 
0.002056. 

First locate the number 20 in the first column and find 
the figure 5 at the top of column, then follow the column 
downward and find the mantissa of log 205 which equals 
0.31 18 and of log 206 which equals 0.3139. 

The difference between these two mantissas is 21 and the 
difference between 2060 and 2050 is 10, also between 2056 
and 2050 is 6. Hence 6/10 of 21 equals 13— or 13. This 
must be added to 0.31 18. This is the mantissa of log 2056, 
which equals 0.313 1. 

Therefore log 2056 = 3.3131 (Ans.). 
log 20.56 = 1.3131 (Ans.). 
log .002056 = 3-3131 (Ans.). 

Example 2: Multiply 20.32 by 0.03849 by 0.0023. 

log 20.32 = i.3079» 
log 0.03849 = 2.5853, 
log 0.0023 = 3-36I7- 



LOGARITHMS 199 

From rule I — log (20.32 X 0.03849 X 0.0023) = 1.3079 

+ 2.5853 + 3-36I7 = 3- 2 549. 

When adding logarithms together, first add the mantissa 
and characteristic separately and then add the two together, 
as this eliminates confusion. That is 

1+2 + 3 = 4. 

0.3079 + 0.5853 + 0.3617 = 1.2549. 

4 + 1.2549 = 3.2549. 

Next, from the same table we find the number whose 
mantissa is 0.2549, that is 0.2549 = log I -799- 

Therefore 3.2549 = log 0.001799 = 0.001799 (Ans.). 
Example 3: Calculate the value (0.023) 5 (using rule No. 3). 

log 0.023 = 2 -36i_7 
I log 0.023 = I X 2.3617 

= I X (2 - 1 + .3617 + 1) 

= |( 3 + 1.3617) = 2 +0.9078 

= 2.9078 

= log 0.08087 = 0.08087 (Ans.). 

EXERCISES 

1. Find the logarithm of 0.049 and 0.49. 

2. Find the logarithm of 3257 and 0.3257. 

3. Find the number of which the logarithm is 3.2833. 

4. Find the number of which the logarithm is 1.9727. 

5. Find the number of which the logarithm is 3.5123. 

6. Find the number of which the logarithm is 0.6395. 

7. Multiply 3.891 by 0.00876. 

8. Multiply 2.562 by 0.01035. 

9. Multiply 0.0323 X 0.00452 X 8890000. 

10. Divide 3320. by 954. 

11. Divide 763 by 0.368. 

12. Divide 316.6 by 4.21. 

13. Calculate 6190 X 3.25 -r- 2625. 

14. Calculate 6542 -i- 318 -5- 66.82. 

15. Find the square root of 3256. and 359200. 

16. Find the cube root of 54.22 and 66.15. 



200 INDUSTRIAL MATHEMATICS 

17. Find the square of 8.252 and 786.1. 

18. Find the fifth power of 0.2382. 

19. Calculate (8o25) 2/s . 



_ , , Vo. 00052 
20. Calculate 



\o. 006814 



Heat 



Heat. — The molecules of all matter are always in motion, 
the rapidity of this motion determines the intensity of heat. 
Heat is the most common form of energy. 

Temperature is the measure of the intensity of molecular 
motion or of the heat, being registered by thermometers or 
pyrometers. There are 2 kinds of thermometer scales in 
general use; the Fahrenheit (F.) and the Centigrade (C). 

The following formulas are used for converting tempera- 
tures given in anyone of the scales to the other scale. 

F.° = (9/5 X C.°) + 32. 
C° = 5/9 X (F.° - 32). 

The Freezing Point of water is 32 ° Fahrenheit, or o° 
Centigrade, and the Boiling Point of water is 212 Fahren- 
heit, or ioo° Centigrade, at atmospheric pressure (14.7 lbs.). 

Absolute Temperature. — The volume of a perfect gas 
increases 1/273 of its volume at o° C. for every increase of 
temperature of i Q C, and decreases 1/273 of i ts volume for 
every decrease of temperature of i° C. Thus, the volume 
of the imaginary gas would be reduced to nothing at — 273 ° 
C. (or — 492 ° F.). This point is called the Absolute zero. 
Absolute temperatures are measured either by the Fahren- 
heit or Centigrade scales, from this zero point. The freezing 
point corresponds to 492 F., or 273 F. absolute. 

Absolute temperature = 460 + F.° 
= 273° + C.° 
Specific Heat of a body is its capacity for heat, or the 
amount of heat in. thermal units required to raise the tern- 



HEAT 201 

perature of I lb. of the body through i degree F., compared 
with that required to raise an equal weight of water i degree. 
If a body with a given temperature is put in a vessel con- 
taining a measured weight of water at a certain temperature, 
the final temperature of the mixture can be found by the 
following formula. 

Heat lost by body = Heat gained by water. 
Weight of body X specific heat X fall of temp. = weight of 
water X specific heat of water or i X rise of temp. 
Example: Find the specific heat of copper from the fol- 
lowing data: § lb. of copper is heated to 212 F. and plunged 
into 20 oz. of water at 6o° F., if the resulting temperature 
was 65. 56 F. 

20 X 1 X (65.56 - 60) 



16 

— X (212 - 65.56) 
2 



0.095 (Ans.). 



Water at 39. i° F 1.00 Steel 0.117 

" " 212 F 1.03 Copper 0.095 

Ice " 32 F 0.504 Charcoal 0.241 

Steam " 212 F 0.481 Air at constant pressure. .0.238 

Mercury 0.033 Oxygen " " . .0.218 

Cast iron 0.13 Hydrogen " " ..3.409 

Wrought iron 0.113 Nitrogen " " ..0.244 

The specific heat of various substances are given in the 
following table. 

A BRITISH THERMAL UNIT (B.t.u.) is the quantity 
of heat required to raise the temperature of 1 lb. of pure 
water i° F. at or near 6o°, at which time water is at its maxi- 
mum density. 

A FRENCH THERMAL UNIT (Calorie) is the quantity 
of heat required to raise the temperature of 1 kilogram 
of pure water i° C. at 15 C. 

1 B.t.u. = 0.252 Calorie. 
1 Calorie = 3.968 B.t.u. 



202 



INDUSTRIAL MATHEMATICS 



The Mechanical Equivalent of Heat is the number of ft. 
lbs. of mechanical energy equivalent to I B.t.u. and is equal 

to 778. 

1 B.t.u. = 778 ft. lbs. 

Latent Heat is the quantity of heat units absorbed or 
given out, in changing one pound of a substance from one 
state to another without changing its temperature. 

When a body passes from the solid to the liquid state, its 
temperature remains stationary at a certain melting point 
during the whole operation of melting and in order to make 
that operation go on, a quantity of heat must be supplied 
to the substance. This quantity of heat is called the Latent 
Heat of Fusion. 

When a body passes from the liquid to the solid state, its 
temperature remains stationary during the whole operation 
of freezing, and a quantity of heat equal to the latent heat 
of fusion is produced in the body and rejected into the 
atmosphere or other surrounding substances. 

When a body passes from the solid or liquid state to the 
gaseous state, its temperature during the operation remains 
stationary at a certain boiling point, depending upon the 
pressure of the vapor produced, and in order to make the 
evaporation go on, a quantity of heat must be supplied to 
the substance evaporated, whose amount for each unit of 
weight of the substance evaporated depends upon the 
temperature. This heat is called the Latent Heat of Evap- 
oration. 

The following table shows the latent heat of various 
substances. 



Latent Heat of Fusion 




Latent Heat of Evaporation 
at Atmospheric Pressure (14.7) 


Material 


B.T.U. 


Material 


B.T.U. 




Boiling 
Point 


B.T.U. 


Ice 


144 
41 
60 


Tin. .. 
Lead . . 
Zinc . . 


26 
10 
5i 


Water. . . 
Alcohol. . 
Ether . . . 


212° F. 

I72°F. 

95° F. 


966 
364 
163 


Cast iron (gray) 

Cast iron (white) .... 



HEAT 203 

Example: Find the B.t.u. required to change 100 lbs. of 
water at ioo° F. into steam at 212 F. 

No. of heat units required to raise = (212-100) X 100 
the temp, of water from ioo° to r= 11,200 B.t.u. 

212°. J 

No. of heat units required to evap- I = 966 X 100 

orate the water from and at 212 . J = 96,600 B.t.u. 

Total heat units = 11,200 + 96,600 = 107,800 B.t.u. (Ans.). 

Expansion of gas: 
1st law — The volume of a given portion of gas varies in- 
versely as its pressure if the temperature is constant. 

P = pressure of gas. 
V — volume of gas. 
PV = PiVi. 

2d law — The increase in volume of a given portion of gas 
varies directly as the increase in temperature, if the 
pressure is constant. 

a = co-efficient of cubical expansion. 

V — initial volume. 
Vi = increase of volume. 
V2 = total volume after increase. 

/ = rise of temp, in degrees. 
V x = Vat. 
V 2 = V x + V = V + Vat = 7(i + at). 

Isothermal expansion or compression takes place when a 
gas is expanded or compressed with an addition or trans- 
mission of sufficient heat to maintain a constant temperature. 

When an expanding gas forces a piston forward against a 
resistance, it does work requiring expenditure of heat. Such 
heat being abstracted from the gas, decreases its temperature. 

Adiabatic expansion or compression takes place when a gas 
is expanded or compressed without the transmission of heat 
15 



where 



204 INDUSTRIAL MATHEMATICS 

to or from it. For example, a gas will follow the adiabatic 
curve if a gas could be expanded or compressed in a receptacle 
of an absolute non-conducting material. 

For the adiabatic expansion of a perfect gas, the following 
formula is used: 

PV r = P 1 V 1 r . 
r = 1.408 for air. 
r — 1.3 for superheated steam. 
r = 10/9 for saturated steam. 
r = 1. 41 1 for carbon monoxide. 
P = initial pressure. 
. Pi = final pressure. 
V = initial volume. 
Vi = final volume. 

EXERCISES 

1. Transfer the following Centigrade reading into Fahrenheit: 

(a) 23 (b) 7 (c) -15° 

2. Transfer the following Fahrenheit reading into Centigrade: 

(a) 63 (b) 15 (c) -4 

3. Change the following temperatures into absolute temperatures: 

(a) 212 F. (6) -6o°F. (c) 12 C. 

4. A vessel containing 200 lbs. of water melts a piece of ice, and the 
fall of temperature is 15 F. What is the weight of the ice? 

5. Equal weight of hot water and ice are placed in one vessel and 
the temperature of water after the ice is melted is 45 ° F. What is the 
initial temperature of hot water? 

6. How many lbs. of hot water at 200 F. will be required to warm 
up a copper plate weighing 40 lbs., from 50 to 150 ? 

7. How many thermal units will be given out in cooling and freezing 
10 lbs. of water at ioo° F.? 

8. How many gallons of gasoline containing a heating power of 
19,000 B.t.u. per lb. will be required in one hour, to develop 10 i.h.p. 
in a gasoline engine, if the thermal efficiency is 17%? (1 lb. =0.17 
gallon.) 

9. What horse power can be developed by using 20 gallons of gaso- 
line per hour, which contains a heating value of 110,000 B.t.u. per gallon, 
if the loss of heat by jacket water is 50%, by exhaust 17% and by radi. 
ation 16% ? 



METAL CUTTING 205 

10. A ioo h.p. steam engine consumes 25 lbs. of steam per i.h.p. per 
hour. How many lbs. of coal will be required in one hour, if the feed 
water temperature is 212 F., and the heating value of coal is 14,000 
B.t.u. per lb.? 

11. Find the compression pressure of a gasoline engine when the 
initial pressure is 13 lbs. per sq. in. and the compression ratio is 4 to 1, 
assuming the expansion of gas follows the adiabatic expansion of super- 
heated steam. The compression ratio is the ratio of the total cylinder 
volume, i.e., compression volume plus piston displacement, to the com- 
pression volume. 

Metal Cutting 

Steel used for cutting metals are broadly classified into 
tool steels and high speed steels. 

Tool Steels or high carbon steels contain 0.60 to 1.50% 
of carbon. The percentage of carbon determines the hard- 
ness of the cutting tool, i.e., the heat treatment being equal. 

High Speed Steels contain several other ingredients such 
as tungsten, chromium, manganese, silicon, molybdenum, 
vanadium and nickel. Of all these ingredients, tungsten 
and chromium are the most important factors as they give 
the steel the property of red hardness; i.e., the tool does 
not lose its cutting ability under very high speeds or heat. 

Cutting Speeds depend upon the following conditions: 

1. Kind of steel to be used, whether tool steel or high speed 

steel. 

2. Shape of tool, whether narrow or broadnosed. 

3. Lip angle of tool, or included angle of nose. 

4. Sharpness of tool. 

5. Position of tool in the tool post. 

6. Depth of cut and amount of feed. 

7. Material to be cut, whether soft, medium or hard, or 

whether brass, cast iron or steel. 

8. Cooling medium, whether used or not, the amount of 

cooling and lubricating effect produced. 

9. Heat treatment of steel. 



206 INDUSTRIAL MATHEMATICS 

io. Elasticity of work or tool, which causes chattering. 
ii. Rigidness with which work is held. 
12. Condition of machine to be used. 

The power required to remove a given amount of metal 
depends upon the shape and sharpness of the cutting tool, 
hardness of the work, depth and feed of cut, lubrication of 
cutting point, and also upon the kind and condition of 
machine. 

The Average Horse Power Required to drive the machine 
can be determined by the product of the amount of chips 
(W) multiplied by 2 constants (Y -\- Z). (Y) varies with 
the kind of material to be cut and (Z) with the kind of ma- 
chine to be used. 

h.p. = Horse power required to drive the machine. 
W = Weight of metal removed per hour in lbs. 
Y = Constant i.o for cast iron. 

1.3 for mild steel. 

2.0 for tool steel. 

0.7 for bronze. 
Z = Constant 0.035 for lathe. 

0.030 for shaper. 

0.025 for miller. 

0.030 for drill, 
h.p. = YZW. 

Example 1 : What horse power will be required to run a 
lathe at 50 r.p.m. to turn a cast iron wheel 8" in diameter, 
with a 1/32" feed and a 1/8" depth of cut? 

Cutting speed = ?r X D X N = 3.1416 X 8 X 50 X 60 

= 753984 in- per hr. 
W = 753984 X d X t X 0.26 

= 75398.4 X 1/32 X 1/8 = 70.686 lbs. 
h.p. = YZW = 1 X 0.035 X 70.686 

= 2.474 (Ans.). 



METAL CUTTING 



207 



The Average Cutting Force at the edge of tool can be 
found by the following formula: 

33000 YZW 28050 YZW 

F = cutting force. 

S = cutting speed in ft. per min. 



X = Cutting Angle 




Taylor Standard 
Cutting Contours 




Back Rake & Clearance 
for Medium Steel & Iron 

Standard Lathe Tool 




^ ^ 



Side Slope for Medium 
Steel and Iron 



Example 2: What is the pressure exerted upon the cutting 
tool in Example 3? 

1, 603,000 FZET 28,050 X 2.474 

W = = = 664.3 lbs. (Ans.). 

5 753984 

12 X 60 

The proper Cutting Speed of a lathe with modern high 
speed tools, can be found by using the following formula: 

7- 



V = cutting speed in feet per minute. 
D = depth of cut, taking 1/64" as a unit. 



208 



INDUSTRIAL MATHEMATICS 



F = feed, taking 1/64" per revolution as a unit. 
H = constant for hardness of material to be cut. 

Hard cast iron or steel 0.6. 
Medium cast iron or steel 1.0. 
Soft cast iron or steel 2.0. 
5 = constant for size of tool: 

sq 



232 for f " 
215 forj" sq. 
325 for f " 
288 for J" 



sq. 
sq. 



tool on cast iron, 
tool on cast iron, 
tool on steel, 
tool on steel. 



Y = constant: 



3 for |" sq. tool on cast iron. 
8 for §" sq. tool on cast iron. 
— 2 for §" sq. tool on steel. 
o for \" sq. tool on steel. 



Z — constant: 



o for I'' sq. 



3." 

4 



tool on cast iron, 
tool on cast iron, 
tool on steel, 
tool on steel. 



0.3 for J" sq. 
0.3 for f" sq. 
0.5 for §" sq. 

With high carbon tool steel the cutting speed is one half 
of the above amount. 

Example: Find the cutting speed of a §" square high 
speed steel tool in a lathe when the depth of cut is 3/16" and 
the feed per rev. is 1/64 upon a piece of soft steel. 

H = 2.0, 5 = 325, D = 12, 

Y= -2, Z = 0.3, >= 1, 

2 X 325 464 



V = 



(l/is 



12 



)(-Vi - 0.3) 



2.15 X 0.84 

= 360 ft. per min. (Ans.). 



POWER REQUIRED 



209 



The Approximate Horse Power Electric Motor Required to Drive 
Various Types of Machines 



Drill presses 
Sensitive — drills up to f", 

I to I h.p. 

12" to 20" 1 " 

24" to 28" 2 " 

30" to 32" 3 " 

Lathes 

6" to 10" 1 h.p. 

12" to 14" 1 to 2 

16" to 20" 2 to 3 

22" to 27" 3 to 5 



Shapers 

10" to 14" stroke 1 to 2 h. 

16" to 18" " 2 to 3 ' 

20" to 24" " 3 to 5 ' 

30" " 5 to -]\ ' 



Planers 



3 h.p. 



Universal milling machines 

No. 1 1 to 2 h.p. 

" 12 2 to 3 " 

" 2 3 to 5 " 

' 3 5 to 7i " 

' 4 7-2 to 10 " 



24" to 27". . 3 to 5 

30" 5 to 7.1 ' 

36" 10 to 15 ' 

42" 15 to 20 ' 

Gear cutters 

36" Xo" 2 to 3 h. 

48" X 10" 3 to 5 ' 

30" X 12" 5 to 7! ' 

72" X 14" 72 to 10 ' 



Grinders 

8" to 10" wheel ,5 h.p. 

12" to 14" wheel 7§ " 

16" to 20" " 10 " 

EXERCISES 

1. Find the horse power required to drive a lathe that cuts 100 lbs. 
of cast iron chips per hour. 

2. What is the horse power required to drive a milling machine running 
at 100 r.p.m. working on a cast iron casting with 0.040" feed per rev., 
cut being £" X 3"? 

3. What horse power will be required to run a lathe at 40 r.p.m. to 
turn a shaft to 2" in diameter, with a ^" feed, and ys" depth of cut? 

4. What is the horse power required to drill a 1" hole in wrought 
iron, using a 0.010" feed, running at 55 r.p.m. 

5. What is the pressure exerted upon the ram of a shaper taking a 
cut in tool steel \" deep and ^5" feed, cutting at a surface speed of 
10 ft. per min.? 

6. Find the proper cutting speed for a \" sq. high speed steel tool 
in a lathe, with the following conditions: 



210 INDUSTRIAL MATHEMATICS 

material = C.I. of medium hardness 

depth of cut = iV 
feed per rev. = j%" 

7. Also find the cutting speed with a f" sq. carbon steel tool: 

material = hard cast iron 

depth of cut = I" 
feed per rev. = ^j" 

8. What is the proper feed for a \' r sq. high speed steel tool, when 
the depth of cut is 3V on a piece of medium hard steel 8" in diameter, 
running at 120 r.p.m.? 

Force, Work, Energy and Momentum 

Force is anything that tends to change the state of a body, 
whether at rest or in motion. A force means the pull, push 
rub, attraction or repulsion of one body upon another, and 
there is always a simultaneous, equal and opposite force 
called the reaction exerted by the second body upon the first. 

Inertia is the property of a body by virtue of which it 
tends to continue in its present state of rest or motion until 
acted upon by some force. 

The Mass of a body is the amount of matter it contains. 
It is equal to its weight divided by the attraction of gravity 
at that particular point. 

TT7 M = mass. 

W 
M = — , where W = weight. 

* g = attraction of gravity. 

Momentum is a term used to designate the product of 
the mass of the body and its velocity at any instant. It 
represents the quantity of motion of a body. 

wv W = weight of body in lbs. 

Momentum = M V = , where M = mass of body. 

*» V = velocity in ft. per sec. 

The rate of change of momentum is proportional to the force 
applied. 



FORCE, WORK, ENERGY AND MOMENTUM 211 

The rate of change of velocity or Acceleration of a body 
caused by a force is proportional to the force applied and 
inversely proportional to its mass. 

F = force in lbs. 

M , , a = acceleration in ft. per sec. 

a = — or F=MX a, y 

F per sec. 

M = mass. 

Example I : If a man pushes a truck weighing i| ton with 
a force of 50 lbs. and the resistance of truck is 20 lbs. per ton 
weight, how long will it take to attain the velocity of 8 miles 
per hour? 

Accelerating force = 50 — (20 X i|) = 2olbs. 

F 20 

= 0.21 ft. per sec. per sec. 
M 



32.16 

5280 X 8 

8 miles per hour = = 11.73 ft. per sec. 

60 X 60 

velocity 11.73 tK x 

time = ; — = = 55.9 sec. (Ans.j. 

acceleration 0.21 

Energy is a capacity of doing work. It is of 2 kinds, 
Potential and Kinetic. 

Potential energy is energy possessed by a body by virtue 
of its condition or position. A weight suspended above the 
ground or a body of water held by a dam possesses potential 
energy. Potential energy also exists as chemical energy in 
storage batteries, etc. 

Kinetic energy is the energy possessed by a body by virtue 
of its motion, and is the work it is capable of performing 
against a retarding resistance before being brought to rest. 
A moving body, a flywheel, a current of air, or a falling body, 
all possess kinetic energy. The kinetic energy of a body is 
equal to the work done upon it to bring it from rest to its 
initial velocity. 



212 INDUSTRIAL MATHEMATICS 

Energy exists in several different forms, but the amount ot 
energy in the universe is fixed and unchangeable. It may 
be transformed from one form to another, but none can be 
created or none destroyed. 

Potential Energy is equal to the product of the force 
tending to produce motion and the distance through which 
the body is able to move. 

E P = W X H. 
Ep = potential energy. 
W = weight of body. 
H = height above ground. 

Kinetic energy is obtained by multiplying one-half of its 
mass by the square of its velocity in ft. per sec. 

WV 2 WV 2 
E K = $MV* = = . 

2g 64.32 

Eg = kinetic energy. 
M = mass of body. 
W = weight of body. 
V = velocity of body in ft. per sec. 

Example 2: What is the potential energy of a stone 
weighing 20 lbs. placed upon a roof 30 ft. high? 

E P = W X H = 20 X 30 = 600 ft. lbs. (Ans.). 

Example 3: If a ball weighing 3 lbs. is thrown vertically 
upward with a velocity of 100 ft. per sec, what is the kinetic 
energy possessed at the start? 

WV 2 3 X 10000 
E K = — = = 466.4 ft. lbs. (Ans.). 

2g 64.32 

Work is the overcoming of resistance through space. If no 
movement is produced no work is done, thus a jack screw 
supporting a load does no work unless the screw is turned. 

Work is equal to the product of the force and the space 



FORCE, WORK, ENERGY AND MOMENTUM 213 

through which it acts. The Unit of Work is the ft. lb. or 
the work done by a force of one lb. acting through a distance 
of one ft. 

W = FS. 
F = force in lbs. 
S = distance in feet. 
W = work in ft. lbs. 

Example 4: Find the work done by the charge, on a pro- 
jectile weighing 100 lbs. which leaves the muzzle of a cannon 
with a velocity of 1000 ft. per second. 

The work done by the projectile is equal to the kinetic 
energy possessed at the breech of the cannon. 

WV 2 100 X 1,000,000 

Therefore work done = = 

2g 64.32 

= 1,554,710.8 ft. lbs. (Ans.). 

The energy of a falling body is equal to the weight multi- 
plied by the height through which it falls. 

The Force of a Blow cannot be expressed directly in 
pounds, but it can be expressed by the average force of blow. 

The average force of blow is equal to the number of foot 
pounds divided by the amount of penetration, plus the 
weight of falling body. 

WS 

Average force of blow = F = -f- W. 

d 

W = weight of falling body in lbs. 
S = height in feet. 
d = distance of penetration in ft. 

Example 5: A single acting steam hammer weighing 500 
lbs. falls through a distance of 4 ft. and compresses the work 
\". What is the average force of blow? 

W X S = 500 X 4 = 2000 ft. lbs. 



214 INDUSTRIAL MATHEMATICS 

2000 2000 

F = + W = + 500 = 96,500 lbs. (Ans.). 

d . 11 

-x— 

4 12 

Power is the rate of doing work and is measured by the 

amount of the work divided by the time in which it is done. 

The unit of power is the horse power (h.p.) which is doing 

work at the rate of 550 ft. lbs. in one sec. or 33000 ft. lbs. in 

one minute. 

FS 
P = — . 
t 

F = force in lbs. 

S = distance in ft. 

/ = time in seconds. 

Example 6: A motor truck weighing 7 tons, attains a 
speed of 15 miles per hour from rest in 20 seconds, during 
which it travels 330 ft., the average resistance of the truck 
being 80 lbs. per ton. Find the average horse power used 
to drive the truck. 

velocity 15 X 5280 

acceleration = = 

time 20 X 60 X 60 

= 1.1 ft. per sec. per sec. 

7 X 2000 

accelerating force = Ma = X 1.1 = 478.8 lbs. 

32.16 

resistance = 80 X 7 = 560 lbs. 

total force = 478.8 + 560 = 1038.8 lbs. 

FXS 1038. 8 X 330 . , A , 

h.p. = = = 31.2 h.p. (Ans.). 

t x 550 20 x 550 

EXERCISES 

1. A shell weighing 100 lbs. is fired vertically upward with a velocity 
of 1500 ft. per sec. What is its kinetic energy at the muzzle of the gun? 

2. What is the potential energy of the above shell when it reaches 
its highest point? 



FORCE, SHEAR AND BENDING MOMENT DIAGRAMS 215 

3. A baseball weighing 9 oz. is dropped from the top of the Washing- 
ton Monument, which is 550 ft. high. What is the energy possessed by 
the ball when it strikes the ground? 

4. A railroad train weighing 100 tons is moving at the rate of 30 
miles per hour. What is the momentum? 

5. An elevator has an upward acceleration of 3.216 ft. per sec. 
per sec. What pressure will a man weighing 150 lbs. exert upon the 
floor of the elevator? 

6. A waterfall is 65 ft. high. If 12 tons of water flow over the falls 
in 1 minute, what kw. generator will the falls run? 

7. What horse power would a turbine develop if it received all the 
water flowing at a rate of 3 miles per hour over a dam 25 ft. high, if the 
water flowed 6" deep over a dam which is 20 ft. long? 

8. A motor truck weighing 7 tons is running at a uniform velocity 
of 15 miles per hr. What is the horse power required, if the traction 
resistance of the road is no lbs. per ton? 

9. A train traveling at a speed of 30 miles per hour is brought to 
rest by a uniform resisting force within a distance of 3000 ft. What - is 
the total resisting force in lbs. per ton? 

10. What horse power will be required to run a train weighing 200 
tons at the rate of 20 miles per hour, if the traction resistance is 10 lbs. 
per ton? 

n. A locomotive has a total weight of 40 tons on the driving wheels 
and the coefficient of friction between the wheels and rails is 0.15. 
What is the maximum pull of the train? 

12. A motor car weighing 3000 lbs. coasts down a slope of 1 in 30 at 
the rate of 15 miles per hour. What is the resistance of the load? 
What is the horse power required to ascend a slope at the same speed? 
(Friction being neglected.) 

13. A stone weighing 50 lbs. falls through a distance of 20 ft. and sinks 
into the ground 15" deep. What is the average force of blow? 

14. A drop hammer weighing 500 lbs. falls through a distance of 4 ft. 
and compresses the work \". What is the force of blow? 

15. At what velocity must a body weighing 5 lbs. be moving in order 
to have stored in it 60 ft. lbs. of energy? 

16. A motor truck weighing 1^ tons is traveling at the rate of 20 
miles per hour, pulling a 7 ton trailer. If the traction resistance of the 
road is 50 lbs. per ton, what is the necessary load on the rear axle, if 
the adhesion of the driving wheels is equal to \ of its load. Also what 
horse power motor is required in the truck? 



216 



INDUSTRIAL MATHEMATICS 



Force Diagrams 

The construction of Force Diagrams somewhat simplify 
the calculation of stresses on beams, etc. 

According to one of the laws of motion, every action or 
force has an equal and opposite reaction; thus when a force 
or load is acting downward on a beam the supports have an 
equal upward reaction. 

In a simple beam which is loaded in the center the reaction 
at each support is equal to one-half of the load. However, 
if the force or load is not central or uniformly distributed the 
reaction at the supports will vary. 

Example: In Fig. I a load of 

21 pounds is resolved into two 

parallel components A and B, a 

distance of 7 and 14 feet, re- 

I : ^"^-^^^ spectively, from the point of 

load. Find the magnitude of 
^---s^ ' force at A and B in pounds. 
I : X_— -■" ^^ First draw a vertical line 1- 

21 to represent 21 lbs. in any 
convenient unit of length, as 
shown, each unit representing 
one pound. 

Next choose a convenient point P and connect points 
1 and 21 with same. 

Choose a convenient point a on line A-i and draw line 
a-b parallel to i-P intersecting line F—b. 

Draw line b—c parallel with 21-P starting from point b. 
Join a and c with a straight line. 

Then by drawing the line X-P parallel with a-c it will 
intersect or divide the line 1-2 1 in the same proportion as the 
load F is distributed at points of support A and B. 

The distance from 1 to X measured in the same scale that 
1-21 was laid out, i.e., 14 units (14 lbs.) will be the magni- 
tude of force at support B. 







Fig. I 



FORCE DIAGRAMS 



217 



Example: Construct a force diagram of a bridge 40 feet 
long with a load of 1000 pounds located 10 feet from one end. 
Solution: Fig. II. 




Fig. II 



Fig. Ill 



A force diagram for a simple beam which is not uni- 
formly loaded is as follows: 

Example: Construct a force diagram for a simple beam 
12 feet long, loaded as shown in Fig. III. 

Solution: First draw the beam to some convenient scale, 
sa Y i" P er f° ot and locate the various loads F 1 , F 2 , F 3 , F i 
and P 5 in their proper position with reference to each other. 

Next draw a vertical line 0-5 to some convenient scale, 
making 0-1 represent 40 lbs., 1-2 equal 60 lbs., 2-3 equal 
100 lbs., 3-4 equal 80 lbs. and 4-5 equal 20 lbs. 

As in Fig. I choose a convenient point P and connect 
points o, 1, 2, 3, 4 and 5 to P. 

Then choose a convenient point a on line A-o and con- 
struct the polygon a-b-c-d-e-f and g, these various lines 
being parallel with lines o-P, i-P, 2-P, 3-P, 4-P and 5-P, 
respectively. 



218 



INDUSTRIAL MATHEMATICS 



If the lines a-b and e-f are continued until they intersect 
at I, this will give the position of the line of the resultant 
force F\ i.e., a point where the total load is concentrated 
or a point where the beam would balance. 

Connecting points a and g with a straight line and con- 
structing line X-P parallel to same will divide line 0-5 in 
proportion to the distribution of the loads at point A and B. 

Thus o-X or R when measured off will equal 160 lbs. or 
the reaction at support A and X-5 or R' will equal 140 lbs. 
or the reaction at support B. 




Fig. IV 



The sum of all the reactions is always equal to the sum 
of all the loads. As shown in Fig. Ill the sum of the reac- 
tions, i.e., 160 and 140 lbs., is equal to the total load or 
40 + 60+100 + 80 + 20 lbs. 



FORCE DIAGRAMS 



219 



Q, 



B® 



A, , , , , I ' A, , 
f J 4 5 6 J 6. 9 10 



Example: Construct a force diagram of a locomotive 
which has its weight distributed as shown in Fig. IV; also 
the resultant reactions at points A and B and the distance 
the resultant force will be from A. 

Solution: Fig. IV. 

Example: Construct a force diagram of a beam loaded as 
shown in Fig. V. p 

Solution: First draw the 
beam to some convenient 
scale and locate the various 
loads and points of support 
in their proper position 
with respect to one another. 

Next draw the vertical 
force line 0-3 to some con- 
venient scale making 0-1 
equivalent to 1000 lbs., 1-3 
equals 3000 lbs. and 2-3 
equals 1000 lbs. 

Then connect points o, 
1, 2 and 3 to P which, as 
previously stated, can be located at any convenient position. 

Construct the force diagram line a-b and c parallel to 
lines o-P and 3-P respectively. 

Draw a-d and / parallel with line o-P beginning at a, also 
draw line I-e-c and / parallel with line 3-P intersecting the 
line a-b and c at c. 

A vertical line projected upward from their intersections, 
i.e., from I to F will give the line of the resultant force. 

By drawing a straight line between points d and e (which 
are the points where the lines a-d-I and I-e-c-f intersect 
the vertical lines of support A-d and B-e respectively), and 
drawing line X—P parallel to same, will intersect or divide 
the line 0-3 at X proportional to the reaction at supports 
A and B. 
16 




Fig. V 



+ Shear 


m 


L\\WS\\V^\ 


r^ 


mwvwwv^ 


r^ 


- Shear 



220 INDUSTRIAL MATHEMATICS 

Thus the distance o-X is equivalent to 1800 lbs. or the 
reaction at A and X-3 is equal to 3200 lbs. or the reaction 
at B. 

Shear Diagram 

A Vertical Shear Diagram may be readily constructed 
from a force diagram. 

For instance at the left hand end of Fig. VII there is 

a 40 lb. force tending to force the 

beam downward, while at point A 

there is a reaction or upward force of 

40 lbs. 

These two forces acting in opposite 
Fig. VI . 6 Kr 

directions tend to shear the beam. 

In a simple beam the shear at opposite ends is equivalent 
to the reaction at these points and the greatest positive 
shear is at the left hand end and the greatest negative shear 
is at the right hand end. 

To construct a vertical shear diagram of a beam, as 
shown in Fig. VII, the force diagram should be constructed 
first as previously explained. (See Fig. III.) 

From X draw the horizontal line X-X' called the Shear 
Axis as shown in Fig. VII. 

The shear diagram above the shear line is positive and 
that below the shear line is negative, i.e., tending to produce 
shear by rotating in the clock-wise direction (such as it 
would to the right of the point of support in Fig. VII), is 
called positive shear; and tending to produce shear by 
rotating counterclock-wise (such as would be produced to 
the left of the support) is called negative shear. 

The vertical shear at any point is equal to the reaction 
at the left hand support, less all the magnitude of forces or 
loads on the beam to the left of the force or load in con- 
sideration. 



SHEAR DIAGRAM 



221 



The amount of shear between A and 2' is constant and is 
equal to the reaction at point A or 40 lbs., and a line drawn 
horizontally from o out as far as the 40 lb. weight extends 
(represented by the line o-a) will be the shear line for the 
40 lb. weight. 

At any point between 2' and 4' the shear will be partially 
counterbalanced by the 40 lb. weight and thus the total 
shear between points 2' and 4' will be equal to 160 lbs. — 40 

. a . <® (gflip^ ,@) , <a> b 

V I' V K' 5' V V 8' 9' f Tl' 

•! — [ 




Fig. VII 



lbs. or 120 lbs.; and a line drawn opposite 1 and extending 
between the 40 and 60 lb. weight, as shown by the line b-c, 
will be the shear line for the 60 lb. weight. 



222 



INDUSTRIAL MATHEMATICS 



At any point between 4/ and 5' the shear will be equal to 
200 lb. — (40 -f- 60 or 100 lbs.) equals 100 lbs. and a line 
d-e drawn opposite 2 and underneath 4' and 5' will be the 
shear line for the 100 lb. weight, etc. 

Then a line connecting the remainder of the points f-g-k 
i-j-k and / will be the shear line and the shaded portion 
will be the shear diagram. 

The portion above the line X—X' will be positive and that 
below the line will be negative. 

In reality the shear line will not be a broken line as shown 
but it will be a curved line, taking the mean average as shown 
by the dotted line S-S' . 

Bending Moment 

The bending moments of a beam is the measure of the 
tendency to produce rota- 
tion about a given point, 
caused by an external force. 
It is equal to the algebraic 
sum of the moments of all 
the external forces acting on 
one side of the section of a 
beam. 

Positive (+) bending mo- 
ments are those which tend 
to produce rotation clockwise 
on the section of the beam 
to the left of a given point 
and those that tend to pro- 
duce motion counter-clockwise are Negative ( — ). 

It is therefore evident that positive bending moments 
will produce convexity upward and negative bending mo- 
ments convexity downward, see Fig. VIII. 

If the force or load is measured in pounds and the distance 




Shear 



Fig. VIII 



BENDING MOMENT 



223 



in feet the moment is expressed in foot pounds, but if the 
distance is measured in inches the moment is expressed in 
inch pounds. The latter is most frequently used. 

Bending moments can be quickly and conveniently deter- 
mined graphically by means of funicular polygons. 

In Fig. IV the polygon a-b-c-d-e-f-g-h-i and j is called 
the Bending Moment Diagram, and the bending moment at 
any point of the beam can be found by multiplying the 
depth of the bending moment diagram at the point in con- 
sideration by the distance P'-P, which is the distance 
from the line 0-4 to P measured at right angles. 

Thus the distance between points h-d measured to the 
same scale that 0-4 was originally laid out to, times the 
distance P' P in inches will be the bending moments in inch 
pounds at that particular point. 

If h-d is equal 10,000 lbs. and P'-P is equal to 3.5" drawn 
to 1/120 size, then the bending moments in inch pounds will 
be equal to 10,000 X 3.5 X 120 or 4,200,000 inch pounds. 




Fig. IX 



224 



INDUSTRIAL MATHEMATICS 



At the other points it will be in proportion to the depth 
of the bending moment diagram. 

Example: Construct a force, vertical shear and bending 



ap f-yyj, i , ij i t i, i i, I, i , 
t i ;:' l 4 4 6 7 8 9 ft' II' 12 13 




Fig. X 



BENDING MOMENT 225 

moment diagram of a cantilever beam projecting 10 ft. out 
from a wall and loaded with a 400 lb. weight 3 ft. from the 
wall, a 500 lb. weight 6 ft. from the wall and a 100 lb. weight 
at the extreme end. 

Solution: Fig. IX. 

Example: Construct a force, vertical shear and a bending 
moment diagram of a beam 13 feet long, supported at the 
extreme left and also at a point 3 feet from the right hand end, 
with a load of 600 lbs. 2 feet from the left end, a load of 1200 
lbs. 5 feet from the left end and a load of 1400 lbs. 12 ft from 
the left end. 

Solution: Fig. X. 

The distances 0-1, 1-2 and 2-3 are laid out representing 
600, 1200 and 1400 lbs. respectively to any convenient scale. 

Points 0-1-2 and 3 are connected to point P which has 
been arbitrarily selected at a point directly underneath A-a 
and an equal distance between points 0-3. 

The line o— P is transferred to the polygon and represents 
line a-b-i, i-P represents b-c, 2-P represents c-d, and 3-P 
represents line I-e-d and g, which intersects line c—d at d, 
or a continuation of line i2'-d'. 

The points a and e, {i.e., where the lines a-b-I and I-e-d-g 
intersect the vertical support lines A— a' -a and B-e'-e) are 
then connected by a straight line a-e. 

This line is transferred to X—P and divides the line 0-3 
in proportion to the distribution of the weights on supports 
A and B, which is approximately 800 lbs. at A and 2400 
lbs. at B. 

It should be noted that line a-e intersects line c-d at /. 
This denotes that the moment to the left of/ is positive and 
that to the right of / is negative. 

To make the bending moment diagram clearer it is fre- 
quently transferred to a parallel base line a'-g' and the positive 
bending moment is placed below the line and the negative 



226 INDUSTRIAL MATHEMATICS 

bending moment is placed above the line as shown in the 
upper bending moment diagram. 

In this case it is found that the maximum negative bending 
moment is greater than the maximum positive bending 
moment, and the beam should be so designed that it will 
withstand the maximum moment whether positive or 
negative. 

To construct the shear diagram the shear line x-p-s-v is 
constructed and, as previously stated, the positive shear, 
or that to the right, is placed above the shear line. 

In this case there are two positive shear diagrams, one 
to the right of each support as shown above the shear line 
while there is only one negative shear which is shown below 
the line. 

Note that the reversal of the shear takes place directly 
underneath the maximum bending moment in all cases. 

In Fig. X the two reversals of shear at p and 5 take place 
directly underneath the maximum positive and negative 
bending moments i'-c' and e'h' '. 

Also in Fig. VII point/ or the reversal of the shear takes 
place beneath the maximum bending moment. 

Pendulum 

A Simple Pendulum is an imaginary one consisting of a 
heavy point suspended by a weightless string. 

A Compound Pendulum is a material body, suspended 
from a fixed axis, about which it oscillates or swings by the 
force of gravity. 

The center of oscillation of a compound pendulum is the 
point at which, if all the matter in the pendulum were con- 
centrated there, it would make a simple pendulum which 
would vibrate or oscillate in the same period of time. 

The angle included between the extreme positions of a 



PENDULUM 227 

line drawn from the point of suspension to the center of 
oscillation is called the Angle of Oscillation. 

The Time of Vibration of a pendulum depends on its 
length and the acceleration of gravity at the given latitude 
and elevation above the sea level. The time of vibration of a 
pendulum varies directly as the square root of its length and 
inversely as the square root of the acceleration of gravity 
at the given point. The time of vibration of a pendulum 
may be varied by adding a weight above the point of sus- 
pension which counteracts the lower weight and lengthens 
the time of vibration. 

A pendulum of a given length always vibrates in the same 
time period at a given locality, provided the angle of oscilla- 
tion does not exceed 5 deg. This property of a pendulum 
is called its Isochronism. 

If a weight suspended by a cord revolves at a uniform 
speed along the circumference of a circle in a horizontal 
plane, this weight forms a conical pendulum, and it is held 
in equilibrium by three forces, the tension in the cord, the 
centrifugal force, and the force of gravity. 

To find the time of vibration of a compound pendulum 
reduce it to an equivalent simple pendulum and find its 
time of vibration. 

At New York City, a pendulum which is 39.1017" long 
(3- 2 585 ft.) will vibrate seconds. 

Abbreviations Used in Formulas 

WiW 2 = Weight of balls in lbs. 
Wz = Weight of bar in lbs. 
d = Distance of center of gravity from point of sus- 
pension. 
L = Length of simple pendulum in inches. 
m = Radius of gyration. 
/ = Time in sec. for "«" oscillations. 



228 



INDUSTRIAL MATHEMATICS 



to = Time in sec. for one revolution. 

g = Value of acceleration of gravity. 

v — Velocity in ft. per sec. of center of gravity of 
weight. 

n — No. of single oscillation in "/" seconds. 
n = No. of rev. per min. 

G = Center of gravity of whole weight. 

P = Point of suspension. 

= Center of oscillation. 

C = Center of gyration. 
Lo = Length of arm of conical pendulum in feet. 

r = Radius of ball circle in ft. ~ L sin a. 

h = Distance of the ball circle below the point of sup- 
port in ft. 



Simple 


Pendulum 


n VZ 
6.25 " 


n = 


6.25* 

VZ * 


m 2 

r 

d ' 






I2gt 2 


39. V 2 




ir 2 n 2 


n 2 






16 



C l ! 

0{ *- ! 



Fig. II 



d = 



Compound Pendulum 
diWi + djWi+.dzW* 



PENDULUM 



229 



d l ^W l + d 2 *W 2 + d?W z 



m l 



d 



Wi + W 2 + W 3 
d?Wi + dJW* + d 3 2 TF a 



diWi + d 2 W 2 + <W 3 
Wi + W 2 + JF 8 

dl'Wl + dfW* + d t *W 9 



L = 




^1 + ^2 + ^3 

_ ^i 2 IFi + d 2 2 W 2 + ^3 2 ^3 

<W 2 + d z W* - diTTi 

-J — ® 

! < G4- 

■ ! ?2 



Fig. Ill 




Fig. IV 



Conical Pendulum 

60 
n 
St 



2irr la 60 

t = — = 6.283 \/f = - 

v \ h n 




230 INDUSTRIAL MATHEMATICS 

54.2 54.2 



27T \h 



■\h -\Lq cos a 

gt 2 2933 

h = — = 0.8146/0 2 = — — = L Q cos a. 

47J- 2 71 Q 2 

EXERCISES 

1. What is the length of a simple pendulum to beat seconds where 
the attraction of gravity is 32.16 ft. per sec. per sec? 

2. What is the attraction of gravity at a point where a simple 
pendulum 39.1393" long, beats seconds? 

3. What is the time of vibration of a simple pendulum 3 ft. long, 
when g = 32.16? 

4. If a simple pendulum is 48" long and g = 32.16, how many beats 
would it make in 20 seconds? 

5. If a compound pendulum is composed of a brass disk \" thick 
and 6" in diameter, the rod is \" in diameter and 66" long, what is the 
time of one beat if g — 32.16? 

6. A compound pendulum consists of 2 cast iron balls 3" and 6" in 
diameter and hangs on a \" steel rod 46.8" long. The small weight 
is 24.3" from the point of support. What is the time of one beat if 
g = 32.16"? 

7. What is the difference in the value of "g" between two points 
A and B, if a pendulum to beat seconds is 39.0152" long at A and is 
39.1393" long at B? 

8. A round steel bar f" in diameter and 5' long is pivoted at a point 
2/5 of its length from one end. On the short end is a cast iron ball 4" 
m diameter and on the long end is a cast iron ball 7" in diameter. If 
the rod just extends through the balls and g = 32.16, what is the time 
of the pendulum? 

9. In a conical pendulum the angle a = 45 deg. If the centers of 
the balls are 6" from the pivot point, how many revolutions per minute 
is the pendulum making? 

10. An engine governor has arms which are 7" long and is turning at 
the rate of 80 revolutions per min.; find the time of one revolution; the 
radius of the ball circle; the distance of the ball circle below the point 
of support; and the angle at which the arms are standing. 




Cam Cutting 



CAM DESIGN 23 1 

Cam Design 

A cam is a mechanical device for converting rotary motion 
into reciprocating motion. Cams are made in several forms 
but usually consist of an irregular shaped disk or of a groove 
cut in a flat or curved surface. 

Cams may be classified in two general classes: according 
to their shape, and also as to the motion they produce. As 
regards shape, we have heart cams, disk cams, face or plate 
cams, and barrel cams. As regards motion, we have uni- 
form acceleration cams, harmonic or crank cams and inter- 
mittent cams. 

A heart cam is the simplest form of cam, and converts 
rotary motion into uniform reciprocating motion. Disk 
cams may have any shape and generally act directly on the 
follower, depending on gravity or a spring to effect the return 
of the follower. Face or plate cams have an irregular groove 
in which the follower moves, cut on the face or plate. They 
may produce any form of motion and have a positive return. 
Barrel cams are cams having a slot cut on the outside of a 
cylindrical surface and may produce any motion. 

A Uniform Velocity Cam is one in which the follower is 
made to pass over equal spaces in equal lengths of time, or 
is impelled from rest to maximum velocity instantly and 
then brought to rest from maximum velocity instantly. 

A Uniform Acceleration Cam or gravity cam is one in 
which the follower is brought from rest to a maximum 
velocity with a uniform acceleration and then brought to 
rest again with a uniform retardation. Since the move- 
ment of the follower is similar to that of a falling body, this 
is also called a gravity curve cam. A Harmonic Cam is one 
in which the follower is brought gradually from rest to 
maximum velocity and then gradually brought to rest, but 
the acceleration is not uniform. This is also called a crank 



232 INDUSTRIAL MATHEMATICS 

cam since the action of the follower is very similar to that 
of a crank. 

An Intermittent Cam may have any irregular movement. 

Uniform velocity cams can only be used at low speeds 
because the abrupt changes from rest to movement and 
vice versa cause too great a shock to the machine at high 
speeds. 

Harmonic and gravity cams are both used for high speed 
work for these cams produce a very smooth working move- 
ment. The harmonic curve, while easy to design does not 
give as smooth action as the gravity curve, but either gives 
good results. 

Intermittent cams are used wherever the service demands 
such a movement, and generally combine periods of rest and 
periods of movement without regard to any set rule. 

There are three distinct phases to the movement of a 
follower, advance, retreat and dwell or rest, and any com- 
bination of these three which returns the follower to its 
starting point, constitutes a cycle. 

If we consider a movement of the follower in any one 
direction as an advance, then any movement in the opposite 
direction is a retreat. 

When a cam, though rotating, produces no movement of 
the follower, the follower is said to dwell or rest. 

To lay out a heart shaped, uniform velocity cam (Fig. I). 
Follower R is to be given a reciprocating motion equal to 
distance i-I: let X be the center of the cam. With X as 
center, draw semi-circle A—S—i, and extend diameter A-X-i 
to /, making I— J equal the required throw; divide i—I into 
any number of equal parts as i, 2, 3, etc. and divide the 
semi-circle by as many radii equally spaced. With X as 
center and radius X-2, draw an arc intercepting X-B at B, 
with same center and radius X-3, draw an arc intercepting 
X-C at C. Continue this process through points 4, 5. 6, 



CAM DESIGN 



233 



etc. obtaining points D, E, F, etc. The latter are points 
on the required curve. The other half of cam is laid out 
in a like manner. 

Since a pointed follower (Fig. I) would cause excessive 
friction, a roller is sometimes used to reduce the friction. 
The curve of the came must be modified slightly, as the 
curve in Fig. I would not give the proper travel to the roller 









Fig. I 




Fig. II 



follower in Fig. II. It is the path of the center of this 
roller which must be considered, as the position of this 
center regulates the throw. The position of this center 
may be determined by adding to each of the distances 
X-B, X-C, etc. in Fig. I, the radius of the roller R in Fig. II, 
thus obtaining the points B, C, etc. in Fig. II. Using these 
points as centers with radius equal to that of the roller, 
describe arcs. A curve drawn tangent to these arcs is the 
desired curve. 

To design a uniform acceleration cam (Fig. Ill) having a 
throw equal to i-G, and an acceleration of two units per 
unit of time, draw semi-circle A-S-i, and divide the cir- 
cumference into 6 equal parts. Project diameter A—X—i 
17 



234 



INDUSTRIAL MATHEMATICS 



to G making i-G equal to the desired throw. Divide l-G 
into two equal parts as 1-4 and 4.-G, and subdivide these 
parts into three divisions whose lengths are to each other 
as 1, 3, 5. With X as a center, draw arcs from 1, 2, 3, etc. 
to intercept the radii in points A, B, C, etc. which are 
points in the cam curve for a pointed follower. Since the 
follower carries a roller, the radius A-R of the roller must 
be added to lengths of X-B, X-C, X-D, etc. and arcs 
drawn from these points K, L, M, etc. using AR as radius. 
A curve, tangent to these arcs, is the desired cam curve. 
If we divide the half-circle into 8 equal parts, then we must 
divide i-G into 8 parts also, in the proportion of 1, 3, 5, 7, 
7, 5, 3» i- With this cam, a follower starts at R, with velocity 
at zero, reaches its maximum velocity at M, and at P, where 
it reverses, its velocity is again zero, making a quiet easy 
working cam suitable for high speeds. 



B 



x k;f > 



|FC jr-T-r- 
PAVV / 



<\-'\ 



< 



€»!■- 



Fig. IV 



To design a harmonic cam (Fig. IV), follower R is to be 
given a harmonic movement which means that R will be 
brought from rest to maximum velocity with a gradual 
acceleration and then brought to rest at the reversing point 



CAM DESIGN 



235 



with a gradual retardation. This cam differs from the 
gravity cam, in that the acceleration and retardation are 
not uniform but variable. 

Draw semi-circle A—S—i and extend diameter to I, making 
i-J equal to the desired throw. Divide 1-7 into any number 
of parts, as eight, whose lengths shall gradually increase 
from 1 to 5 and gradually decrease from 5 to I. Divide the 
semi-circle by as many radii equally spaced. Draw the arcs 
2-B, 3— C, 4~D, etc., intercepting the radii at points B, C, 
D, etc, A curve passing through these points will be the 
desired curve. For a roller follower, add radius of roller to 
lengths X-B, X-C, X-D, etc. and proceed as in Fig. II. 
This cam can be used for high speeds, although it is not as 
easy working as a gravity curve cam. 

Effect of changing location of cam roller 





Fig. V 



Fig. VI 



When the line of motion of a follower passes through the 
center of the cam, and the angle of the cam causes it to work 
hard, the curve may be modified and the same movement of 
the follower obtained by placing the follower so that its 
line of movement is parallel to its former position, but not 
passing through the center of the cam. As example, in 
Fig. V: Here the cam, rotating in direction of arrow (A) moves 
the follower in direction of arrow (B). The angle of the 
cam with the follower at beginning of stroke is 30 deg. as 
determined by a tangent to the curve. Should the cam 
work hard, this could be remedied by increasing the diameter 



236 INDUSTRIAL MATHEMATICS 

of the cam which would reduce the angle of the cam. Some- 
times this cannot be done owing to the design of the machine, 
but the same result can be accomplished by changing the 
location of the roller. Fig. VI has the same condition as 
Fig. V, but the cam roller has been moved above the line 
of center of the cam. The line of motion now passes along 
line L-M and angle of cam is 20 deg., making an easier 
working cam. The roller must always be offset in the direc- 
tion opposite to rotation of the cam and the angle of the cam 
decreases as the offset increases. However, if the follower 
be offset too much, the thrust at right angles to line of motion 
will increase the friction until the good effects of the offset 

are overcome. 

EXERCISES 

Note. — (Draw all cams on a 4" diameter base circle.) 

1. Lay out a heart-shaped, uniform velocity cam for a pointed 
follower, that will give a reciprocating motion of 2" to the follower. 

2. Lay out a similar cam to give the same movement to a follower 
with a roller if" in diameter. 

3. Lay out a harmonic cam to give a reciprocating movement of 2§" 
to a follower with a roller \\" in diameter. 

4. Draw a gravity curve cam to give a reciprocating movement of 
1 \" to a roller follower: radius of roller \" and an acceleration of 3. 

5. Lay out a uniform velocity cam which shall advance f" during a 
30 deg. revolution of the cam; dwells for 60 deg., retreats f" during 
next 30 deg. and dwells until the end of the cycle. Show (by sketch) 
that this cam will work easier if the roller is offset f " '. 

REVIEW EXERCISES 

1. At what speed in r.p.m. must a 6" grinding wheel run to attain a 
surface speed of 6000 ft. per minute? 

2. At what surface speed in feet per minute is a 20" grinding wheel 
running if the spindle runs at 1150 r.p.m.? 

3. How many graduations must I have on my dial to represent 
0.001" movement of a tool on a cross slide of a lathe carriage, if an 8 
pitch screw is used? 

4. Find the number of graduations on a dial to read thousandths, 
on an elevating screw of a milling machine, with a reduction of 3 to 4 
in bevel gearing, if a 6 P screw is used. 



REVIEW EXERCISES 237 

5. If the arm ratio of an indicator is § to 11, what error does a ^j" 
movement on the long arm represent? 

6. What pressure is required to punch a hole 2" in diameter in a soft 
steel plate f" thick, if the shearing resistance of steel is 60,000 lbs.? 

7. What weight will a §" round steel rod support, under tension, if 
the steel has a tensile strength of 100,000 lbs. per square inch, providing a 
factor of safety of 8 is used? 

8. A longitudinal steel boiler stay 20 ft. long and 1" in diameter 
supports a flat area of 10" square, having on it a pressure of 120 lbs. per 
sq. inch. Find the greatest stress in the stay due to its own weight and 
the steam pressure. 

9. What must be the diameter of a steel car axle to resist the shearing 
stress of a load of 80,000 lbs., using factor of safety 15. 

10. How many f " studs must be used to hold a 24" cylinder head of a 
steam engine, if the maximum steam pressure is 125 lbs. per sq. in., 
allowing a factor of safety of 8. The material from which the studs are 
made has a tensile strength of 60,000 lbs. per sq. inch. Use the root 
diameter of the studs as the effective area. 



Decimal Equivalents, 



APPENDIX 

TABLE I 
Squares and Square Roots of Fractions 



Fraction 


Decimal 
Equivalent 


Square 


Sq. Root 






1/64 


-0.015625 


0.000244 


0.1250 




1/32 




—0.03125 


0.000977 


0.1768 






3/64 


—0.046875 


0.0022 


0.2165 


1/16 






—0.0625 


0.0039 


0.2500 






5/64 


—0.078125 


0.0061 


0.279s 




3/32 




-0.09375 


0.0088 


0.3062 






7/64 


-0.109375 


0.0119 


0.3307 


1/8 






—0.125 


0.0156 


0.3535 






9/64 


—0.140625 


0.0198 


0.3750 




5/32 




—0.15625 


0.0244 


o.3953 






11/64 


-0.171875 


0.0295 


0.4161 


3/i6 






-0.1875 


0.0352 


0.4330 






13/64 


—0.203125 


0.0413 


0.4507 


, 


7/32 




-0.21375 


0.0479 


0.4677 * 






15/64 


-0.234375 


0.0549 


0.4841 


1/4 






—0.250 


0.0625 


0.5000 






17/64 


—0.265625 


0.0706 


0.5154 




9/32 




—0.28125 


0.0791 


0.5303 






19/64 


—0.296875 


0.0881 


0.5449 


5/i6 






-0.3125 


0.0977 


0.5590 






21/64 


—0.328125 


0.1077 


0.5728 




n/32 




-0.34375 


0.1182 


0.5863 






23/64 


-0.359375 


0.1291 


0.5995 


3/8 






-0.375 


0.1406 


0.6124 






25/64 


—0.390625 


0.1526 


O.6250 




13/32 




—0.40625 


0.1650 


O.6374 






27/64 


—0.421875 


0.1780 


O.6495 


7/16 






-0.4375 


0.1914 


O.6614 






29/64 


-0.453125 


0.2053 


0.6732 




IS/32 




—0.46875 


0.2197 


O.6847 






31/64 


-0.484375 


0.2346 


O.6960 


1/2 






-o.S 


0.2500 


O.7071 






33/64 


-0.515625 


0.2659 


O.7181 




17/32 




-0.53125 


0.2822 


0.7289 






35/64 


-0.54687S 


0.2991 


0.7395 


9/i6 






—0.5625 


0.3164 


0.7500 






37/64 


-0.578125 


0.3342 


O.7603 




19/32 




-0.59375 


0.352s 


O.7706 






39/64 


-0.609375 


0.3713 


O.7806 


5/8 






-0.625 


0.3906 


O.7906 






41/64 


—0.640625 


0.4104 


O.8004 




21/32 




—0.65625 


0.4307 


0.8101 






43/64 


—0.671875 


0.4514 


O.8197 


11/16 






-0.6875 


0.4727 


O.8292 






45/64 


—0.703125 


0-4944 


0.8385 




23/32 




-0.71875 


0.5166 


0.8478 






47/64 


-0.734375 


0-5393 


O.8569 


3/4 






-0.750 


0.5625 


O.8660 






49/64 


—0.765625 


0.5862 


O.8750 




25/32 




—0.78125 


01.6104 


0.8839 






51/64 


-0.796875 


0.6350 


O.8927 


I3/I6 






—0.8125 


0.6602 


O.9OI4 






53/64 


-0.828125 


0.6858 


O.9100 




27/32 




-0.84375 


0.7119 


O.9186 






55/64 


-0.859375 


0.7385 


O.9270 


7/8 






-0.875 


0.7656 


0.9354 






57/64 


—0.890625 


0.7932 


0-9437 




29/32 




—0.90625 


0.8213 


0.9520 






59/64 


—0.921875 


0.8499 


O.9601 


IS/16 






-0.9375 


0.8789 


O.9682 






61/64 


-0.953125 


0.9084 


0.9763 




31/32 




—0.96875 


0.9385 


O.9843 






63/64 


-0.984375 


0.9690 


0.9922 


I 






— 1. 000 


1. 0000 


1.0000 



NATURAL TRIGONOMETRICAL FUNCTIONS 



239 



TABLE II 
Natural Trigonometrical Functions 



D 


M 



Sines 


Cosines 


Tang. 


Cotang. 


JL 


M 





.00000 


1. 0000 


.00000 


Infinite 


90 







15 


.00436 


•99999 


.00436 


229.182 




45 




30 


.00873 


.99996 


.00873 


114-589 




30 




45 


.01309 


•99991 


.01309 


76.3900 




15 


1 





•01745 


.99985 


.01745 


57.2900 


89 







15 


.02181 


.99976 


.02182 


45.8294 




45 




30 


.02618 


.99966 


.02618 


38.1885 




30 




45 


•03054 


•99953 


.03055 


32.7303 




15 


2 





.03490 


•99939 


.03492 


28.6363 


88 







15 


.03926 


.99923 


.03929 


25.4517 




45 




30 


.04362 


.99905 


.04366 


22.9038 




30 




45 


.04798 


.99885 


.04803 


20.8188 




15 


3 





.05234 


.99863 


.05241 


19.0811 


87 







15 


.05669 


.99839 


.05678 


17.6106 




45 




30 


.06105 


.99813 


.06116 


16.3499 




30 




45 


.06540 


.99786 


•06554 


15.2571 




15 


4 





.06976 


.99756 


.06993 


14-3007 


86 







15 


.07411 


•99725 


.07431 


13.4566 




45 




30 


.07846 


.99692 


.07870 


12.7062 




30 




45 


.08281 


.99656 


.08309 


12.0346 




15 


S 





.08716 


.99619 


.08749 


1 1. 430 1 


85 







15 


.09150 


.99580 


.09189 


10.8829 




45 




30 


.09585 


•99540 


.09629 


10.3854 




30 




45 


.10019 


•99497 


.10069 


9-93IOI 




15 


6 





.10453 


•99452 


.10510 


9.5I436 


84 







15 


.10887 


.99406 


.10952 


9.13093 




45 




30 


.11320 


•99357 


.11393 


8.77689 




30 




45 


•II754 


.99307 


.11836 


8.44896 




15 


7 





.12187 


.99255 


.12278 


8.14435 


83 







15 


.12620 


.99200 


.12722 


7.86064 




45 




30 


.13053 


.99144 


•13165 


7-59575 




30 




45 


.13485 


.99086 


.13609 


7.34786 




15 


8 





.13917 


.99027 


•14054 


7.H537 


82 







15 


.14349 


.98965 


.14499 


6.89688 




45 




30 


.14781 


.98902 


•14945 


6.69116 




30 




45 


.15212 


.98836 


.15391 


6.49710 




15 


9 





.15643 


.98769 


.15838 


6.31375 


81 







15 


.16074 


.98700 


.16286 


6.14023 




45 




30 


.16505 


.98629 


.17634 


5.97576 




30 




45 


.16935 


.98556 


.17183 


5.81966 




15 


10 





.17365 


.98481 


.17633 


5.67128 


80 







IS 


•17794 


.98404 


.18083 


5.53007 




45 




30 


.18224 


.98325 


•18534 


5.39552 




30 




45 


.18652 


.98245 


.18986 


5.26715 




15 


11 





.19081 


.98163 


•19438 


5.14455 


79 







15 


.19509 


.98079 


.19891 


5-02734 




45 




30 


•19937 


•97992 


.20345 


4-91516 




30 




45 


.20364 


•97905 


.20800 


4.80769 




15 


12 





.20791 


.97815 


.21256 


4-70463 


78 







15 


.21218 


.97723 


.21712 


4.60572 




45 




30 


.21644 


.97630 


.22169 


4-5I07I 




30 




45 


.22070 


•97534 


.22628 


4.41936 




15 


13 





.22495 


•97437 


.23087 


4-33I48 


77 







15 


.22920 


.97338 


•23547 


4-24685 




45 




30 


•23345 


•97237 


.24008 


4.16530 




30 




45 


.23769 


.97134 


.24470 


4.08666 




IS 


14 





.24192 


•97030 


•24933 


4.01078 


76 







15 


.24615 


.96923 


•25397 


3-93751 




45 




30 


.25038 


.96815 


.25862 


3.86671 




30 




45 


.25460 


.96705 


.26328 


3-79827 




15 


IS 





.25882 


.96593 


.26795 


3.73205 


75 






240 



INDUSTRIAL MATHEMATICS 



TABLE II— (Continued) 



D 


M 


Sines 


Cosines 


Tang, 


Cotang, 


D 


M 


15 





.25882 


.96593 


.26795 


3-73205 


75 







15 


.26303 


.96479 


.27263 


3.66796 




45 




30 


.26724 


■96363 


.27732 


3.60588 




30 




45 


.27144 


.96246 


.28203 


3-54573 




15 


16 





.27564 


.96126 


.28674 


3.48741 


74 







15 


•27983 


.96005 


.29147 


3.43084 




45 




30 


.28402 


.95882 


.29621 


3-37594 




30 




45 


.28820 


•95757 


.30096 


3.32264 




15 


17 





.29237 


.95630 


.30573 


3.27085 


73 







15 


.29654 


•95502 


.31051 


3.22053 




45 




30 


.30070 


•95372 


.31530 


3-I7I59 




30 




45 


.30486 


.95240 


.32010 


3.12400 




15 


18 





.30902 


.95106 


.32492 


3.07768 


72 







15 


.31316 


•94970 


.32975 


3.03260 




45 




30 


.31730 


.94832 


•33459 


2.98868 




30 




45 


.32144 


.94693 


•33945 


2.94591 




15 


19 





.32557 


•94552 


•34433 


2.90421 


71 







15 


.32969 


.94409 


•34921 


2.86356 




45 




30 


.33381 


.94264 


•35412 


2.82391 




30 




45 


•33792 


.94118 


.35904 


2.78523 




15 


20 





.34202 


.93969 


.36397 


2.74748 


70 







15 


.34612 


.93819 


.36892 


2.71062 




45 




30 


.35021 


.93667 


.37388 


2.67462 




30 




45 


.35429 


.93514 


.37887 


2.63945 




IS 


21 





.35837 


.93358 


.38386 


2.60509 


69 







15 


.36244 


.93201 


.38888 


2.57150 




45 




30 


.36650 


.93042 


•39391 


2.53865 




30 




45 


.37056 


.92881 


.39896 


2.50652 




15 


22 





.37461 


.92718 


.40403 


2.47509 


68 







15 


.37865 


.92554 


.40911 


2.44433 




45 




30 


.38268 


.92388 


.41421 


2.41421 




30 




45 


.38671 


.92220 


.41933 


2.38473 




15 


23 





.39073 


.92050 


•42447 


2.35585 


67 







15 


•39474 


.91879 


.42963 


2.32756 




45 




30 


.39875 


.91706 


.43481 


2.29984 




30 




45 


.40275 


•9IS3I 


.44001 


2.27267 




15 


24 





.40674 


.91355 


.44523 


2.24604 


66 







15 


.41072 


.91176 


.45047 


2.21992 




45 




30 


.41469 


.90996 


•45573 


2.19430 




30 




45 


.41866 


.90814 


.46101 


2.16917 




15 


25 





.42262 


.90631 


.46631 


2.14451 


65 







15 


.42657 


.90446 


.47163 


2.12030 




45 




30 


.43051 


.90259 


.47697 


2.09654 




30 




45 


•43445 


.90070 


.48234 


2.07321 




15 


26 





•43837 


.89879 


.48773 


2.05030 


64 







15 


.44229 


.89687 


.49314 


2.02780 




45 




30 


.44620 


.89493 


.49858 


2.00569 




30 




45 


•45010 


.89298 


.50404 


1.98396 




15 


27 





•45399 


.89101 


.50952 


1.96261 


63 







15 


.45787 


.88902 


.51503 


1.94162 




45 




30 


.46175 


.88701 


•52057 


1.92098 




30 




45 


.46561 


.88499 


.52612 


1.90069 




15 


28 





.46947 


.88295 


•53I7I 


1.88073 


62 







15 


•47332 


.88089 


•53732 


1. 86109 




45 




30 


.47716 


.87882 


.54295 


1. 84177 




30 




45 


.48099 


.87673 


.54862 


1.82276 




IS 


29 





.48481 


.87462 


•55431 


1.80405 


61 







15 


.48862 


.87250 


.56003 


1.78563 




45 




30 


.49242 


.87036 


■56577 


1.76749 




30 




45 


.49622 


.86820 


.57155 


1.74964 




15 


30 





.50000 


.86603 


•57735 


1.73205 


60 






NATURAL TRIGONOMETRICAL FUNCTIONS 



24I 



TABLE II— (Continued) 



D 


M 


Sines 


Cosines 


Tang. 


Co tang. 


D 


M 


30 





.50000 


.86603 


•57735 


1.7.3205 


60 







15 


.50377 


.86384 


.58318 


1. 71473 




45 




3'0 


•50754 


.86163 


.58904 


1.69766 




Co 




45 


.51129 


.85941 


•59494 


1.68085 




15 


31 





.51504 


.85717 


.60086 


1.66428 


59 







15 


.51877 


.85491 


.60681 


1.64795 




45 




30 


.52250 


.85264 


.61280 


1.63J8S 




30 




45 


.52621 


.85035 


.61882 


1.61598 




15 


32 





.52992 


.84805 


.62487 


1.60033 


58 







15 


.53361 


.84573 


.63095 


1.58490 




45 




30 


•53730 


.84339 


.63707 


1.56969 




30 




45 


.54097 


.84104 


.64322 


1.55467 




15 


33 





.54464 


.83867 


.64941 


1.53986 


57 







15 


.54829 


.83629 


.65563 


1.52525 




45 




30 


.55194 


.83389 


.66188 


1. 51084 




30 




45 


•55557 


.83147 


.66818 


1. 49661 




15 


34 





.55919 


.82904 


.67451 


1.48256 


56 







15 


.56280 


.82659 


.68087 


1.46870 




45 




30 


.56641 


.82413 


.68728 


1-45501 




30 




45 


.57000 


.82165 


.69372 


1. 44149 




15 


35 





.57358 


.81915 


.70021 


1.42815 


55 







15 


.57715 


.81664 


.70673 


1.41497 




45 




30 


.58070 


.81412 


.71329 


1.40195 




30 




45 


.58425 


.81157 


.71990 


1.38909 




15 


36 





.58779 


.80902 


.72654 


1.37638 


54 







15 


.59131 


.80644 


•73323 


1.36383 




45 




30 


.59482 


.80386 


.73996 


I.35I42 




30 




45 


.59832 


.80125 


.74673 


I.339I6 




15 


37 





.60181 


.79864 


•75355 


1.32704 


53 







15 


.60529 


.79600 


.76042 


I.3I507 




45 




30 


.60876 


•79335 


.76733 


1.30323 




30 




45 


.61222 


.79069 


.77428 


1. 29152 




15 


38 





.61566 


.78801 


.78129 


1.27994 


52 







15 


.61909 


.78532 


.78834 


1.26849 




45 




30 


.62251 


.78261 


•79543 


1. 25717 




30 




45 


.62592 


.77988 


.80258 


1.24597 




15 


39 





.62932 


.77715 


.80978 


1.23490 


51 







15 


.63271 


•77439 


.81703 


1.22394 




45 




30 


.63608 


.77162 


.82434 


1.21310 




30 




45 


•63944 


.76884 


.83169 


1.20237 




15 


40 





.64279 


.76604 


.83910 


I.I9I75 


50 







15 


.64612 


.76323 


.84656 


1.18125 




45 




30 


.64945 


.76041 


.85408 


1.17O85 




30 




45 


.65276 


.75756 


.86165 


1.160S6 




15 


41 





.65606 


•75471 


.86929 


1. 15037 


49 







15 


.65935 


.75184 


.87698 


1. 14029 




45 




30 


.66262 


.74896 


.88472 


1. 13029 




30 




45 


.66588 


.74606 


.89253 


1. 1 2041 




15 


42 





.66913 


.74314 


.90040 


1.11061 


48 







15 


.67237 


.74022 


.90834 


1.10091 




45 




30 


■67559 


.73728 


.91633 


1.09131 




30 




45 


.67880 


•73432 


.92439 


1. 08179 




15 


43 





.68200 


•73135 


.93251 


1.07237 


47 







15 


.68518 


.72837 


.94071 


1.06303 




45 




30 


.68835 


.72537 


.94896 


1.05378 




30 




45 


.69151 


.72236 


.95729 


1. 04461 




15 


44 





.69466 


.71934 


.96569 


1.03553 


46 







15 


.69779 


.71630 


.97416 


1.02653 




45 




30 


.70091 


.71325 


.98270 


1.01761 




30 




45 


.70401 


.71019 


.99131 


1.00876 




15 


45 





.70711 


.70711 


1. 00000 


1. 00000 


45 






242 



INDUSTRIAL MATHEMATICS 



TABLE III 
Common Logarithms 



N 





I 


2 


3 


4 


5 


6 


7 


8 


9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


ii 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1 106 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


356o 


3579 


3598 


23 


3617 


3636 


3655 


3674 


3692 


37ii 


3729 


3747 


3766 


3784 


24 


3802 


'3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


S038 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


35 


5441 


5453 


5405 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


36 


5563 


5575 


5587 


5599 


561 1 


5623 


5635 


5647 


5658 


5670 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


39 


5911 


5922 


5933 


5944 


5955 


'5966 


5977 


5988 


5999 


6010 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


4i 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


43 


6335 


6345 


6355 


6365 


6375 


638S 


6395 


6405 


64LS 


6425 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


49 


6902 


691 1 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


73i6 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


59 


7709 


77i6 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 



COMMON LOGARITHMS 



243 



TABLE III— (Continued) 



N 


O 


1 


2 


3 


4 


5 


6 


7 


8 


9 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 


9085 


9090 


9096 


9IOI 


9106 


9112 


9117 


9122 


9128 


9133 


82 


9138 


9143 


9149 


9154 


9159 


916S 


9170 


9175 


9180 


9186 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


86 


9345 


9350 


9355 


936o 


9365 


9370 


9375 


9380 


9385 


9390 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


958i 


9586 


9i 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


99 


9956 


996i 


996s 


9969 


9974 


9978 


9983 


9987 


9991 


9996 



Specific Gravity of Materials 
TABLE IV 

Water 1.00 Copper 

Kerosene 0.80 Silver 

Oak 0.80 Lead 

Gasoline 0.90 . Mercury 

Aluminum . 2.58 Tungsten 

Tin 7.29 Gold 

Iron 7.40 Platinum , 



8.90 
10.50 
11.30 
13-60 
18.80 
19.30 
21.50 



244 



INDUSTRIAL MATHEMATICS 



TABLE V 
Weight and Specific Gravity of Liquids 



Liquids at 32 F. 



Weight of 
1 Cu. Ft. 
in Lbs. 



Weight of 

1 Gal. (Imp.) 

in Lbs. 



Specific 

Gravity 

Water = u 



Mercury 

Bromine 

Sulphuric acid 

Nitrous acid 

Chloroform 

Water of the Dead Sea 

Nitric acid 

Acetic acid 

Milk 

Sea water 

Pure water (distilled) at 

39° F 

Oil, linseed 

Oil, whale 

Oil, turpentine 

Petroleum 

Naphtha 

Ether, nitric 

Ether, sulphurous 

Ether, acetic 

Ether, hydrochloric 

Ether, sulphuric 

Alcohol, proof spirit 

Alcohol, pure 

Benzine 



848.7 
185. 1 
1 14.9 

96.8 

95-5 

77-4 

76.2 

67.4 

64-3 

64.05 

62.425 

58.7 

57.4 

54-3 

54-9 

53-1 

69-3 

67.4 

55-6 

54-3 

44.9 

57.4 
49-3 
53-1 



136.0 
29.7 
18.4 
15-5 
15-3 
12.4 
12.2 
10.8 
10.3 
10.3 

10. 
9.4 
9.2 
8.7 
8.8 

8-5 

11. 1 
10.8 

8.9 
8.7 
7.2 
9.2 
7-9 
8.5 



I3.596 
2.966 
1.84 
i-55 
1-53 
1.24 
1.22 
1.08 
1.03 
1.026 

1.0 

0.94 
0.92 
0.87 
0.88 
0.85 
1. 11 
1.08 
0.89 
0.87 
0.72 
0.92 
0.79 
0.85 



TABLE VI 
Melting Point of Materials 



Mercury 38 

Tin 450 

Lead 621 

Zinc 787 

Antimony 1166 

Aluminum 12 18 

Radium 1 292 

Barium 1562 

Bronze 1675 

Silver 1762 

Gold 1945 

Copper 198 1 

Cast iron 2300 



Manganese 2300° 

Steel 2500 

Silicon 2588 

Nickel 2646 

Cobalt 2696 

Wrought iron 2900 

Vanadium 3191 

Platinum 3200 

Titanium 3272 

Uranium 3362 

Molybdenum 4500 

Tungsten 5430 

Carbon 6500 



TAPERS AND ANGLES 



245 



TABLE VII 
Strength of Miscellaneous Metals 



Material 


Modulus of 
Rigidity 


Elastic 
Limit 


Tensile 

Strength 


Phosphor bronze 


13,000,000 
15,000,000 
15,000,000 
28,000,000 
28,000,000 
29,000,000 
28,000,000 
30,000,000 
30,000,000 


20,000 
35.000 
50,000 
30,000 
40,000 
80,000 
80,000 
160,000 
220,000 


30,000 
50,000 
75,000 
40,000 
60,000 
110,000 


Manganese bronze 


Aluminum bronze 




Mild open hearth steel 

Tool steel 


Nickel steel 


110,000 


Nickel chrome steel 

Vanadium steel treated 


180,000 
228,000 



TABLE VIII 
Tapers and Angles 





Included 


With Center 




Taper per 
Inch from 


Taper 


<. 


Line ' 


Taper 
per Inch 


per Foot 














Deg. 


Min. 


Deg. 


Min. 






1 

8 





36 





18 


.010416 


.005203 


16 





54 





27 


.015625 


.007812 


1 
4 


1 


12 


O 


36 


.020833 


.010416 


16 


I 


30 


O 


45 


.026042 


.013021 


8 


1 


47 





53 


.031250 


.015625 


16 


2 


05 




02 


.036458 


.018229 


1 
2 


2 


23 




11 


.041667 


.020833 


£ 


2 


42 




21 


.046875 


.023438 


8 


3 


00 




30 


.052084 


.026042 


16 


3 


18 




39 


.057292 


.028646 


4 


3 


25 




47 


.062500 


.031250 


13 
T6 


3 


52 




56 


.067708 


.033854 


8 


4 


12 


2 


06 


.072917 


.036456 


15. 
T6 


4 


28 


2 


14 


.078125 


.039063 


I 


4 


45 


2 


23 


.083330 


.041667 


ii 


5 


58 


2 


59 


.104666 


.052084 


i* 


7 


08 


3 


34 


.125000 


.062500 


if 


8 


20 


4 


10 


.145833 


.072917 


2 


9 


32 


4 


46 


.166666 


.083332 


a* 


11 


54 


5 


57 


.208333 


.104166 


3 


14 


16 


7 


08 


.250000 


.125000 


3i 


16 


36 


8 


18 


.291666 


•145833 


4 


18 


54 


9 


27 


•333333 


.166666 


A\ 


21 


40 


10 


50 


.375000 


.187500 


5 


24 


04 


12 


02 


.416666 


.208333 


6 


28 


06 


14 


03 


.500000 


.250000 



246 



INDUSTRIAL MATHEMATICS 



Brown and Sharp Standard Tapers 

No. of Taper, 4 5 7 9 

Diameter at small end, .35 in. .45 in. .60 in. .90 in. 

This taper is \" per foot 



No. of Taper, 
Diameter at small end, .37 in. 

This taper is about f " per foot. 



Morse Standard Tapers 

1234 

.57 in. .78 in. 1.02 in, 



TABLE IX 
Table of Speeds 









Cutting Speeds in Feet per Minute 






Diam. 


20 


30 


40 


50 


60 


70 


80 


90 


100 


Ins. 








Revolutions per 


Minute 








1 
4 


306 


458 


611 


764 


916 


1070 


1222 


1376 


1528 


Is 


204 


306 


407 


509 


612 


712 


814 


916 


1019 


1 
2 


153 


229 


306 


382 


458 


534 


612 


688 


764 


5 

8 


122 


183 


244 


306 


366 


428 


488 


550 


611 


3 

4 


102 


153 


204 


255 


306 


356 


408 


458 


509 


7 
8 


87 


131 


175 


218 


262 


306 


350 


392 


437 


I 


76 


115 


153 


191 


230 


268 


306 


344 


382 


I* 


68 


102 


136 


170 


204 


238 


272 


306 


340 


it 


61 


92 


122 


153 


184 


214 


244 


274 


306 


if 


56 


83 


III 


139 


167 


194 


222 


250 


278 


I* 


51 


7 6 


102 


127 


152 


178 


204 


228 


255 


If 


47 


71 


94 


118 


141 


165 


188 


212 


235 


if 


44 


65 


87 


109 


130 


152 


174 


196 


218 


ll 


4i 


6l 


82 


102 


122 


143 


163 


183 


204 


2 


38 


57 


76 


95 


114 


134 


152 


172 


191 


2* 


36 


54 


72 


90 


108 


126 


144 


162 


180 


2| 


34 


5i 


68 


85 


102 


119 


136 


153 


170 


2f 


32 


48 


64 


80 


97 


112 


129 


145 


161 


2\ 


3i 


46 


61 


76 


92 


106 


122 


134 


153 


2 8 


29 


44 


58 


73 


88 


102 


117 


130 


146 


3* 


28 


42 


56 


70 


83 


97 


in 


125 


139 


af 


27 


40 


53 


67 


80 


93 


106 


119 


133 


3 


25 


38 


5i 


64 


76 


90 


102 


114 


127 



CUTTING SPEEDS 



247 







Cutting Speeds 


n Feet per Minute 






Diam. 


no 


120 


130 


140 


ISO 


160 


no 


180 


Inches 






Revolutions 


per Minute 






1 


1681 


1833 


1986 


2139 


2292 


2462 


2615 


2780 


3 

8 


II20 


1222 


1324 


1426 


1528 


1632 


1735 


1836 


1 

2 


84O 


917 


993 


1070 


1 146 


1221 


1298 


1374 


5 

8 


672 


733 


794 


856 


917 


976 


IO36 


1098 


3 
4 


560 


611 


662 


713 


764 


816 


867 


918 


7 
8 


480 


524 


568 


611 


655 


699 


742 


786 




420 


458 


497 


535 


573 


611 


649 


687 


1 S 


373 


407 


441 


475 


509 


542 


576 


610 


T 1 
^4 


336 


367 


397 


428 


458 


489 


520 


551 


T 3 
*8 


306 


333 


361 


389 


417 


444 


472 


500 


T l 

l 2 


280 


306 


331 


357 


382 


407 


433 


458 


T 5 

1 8" 


259 


282 


306 


329 


353 


377 


400 


423 


T 3 


246 


262 


284 


306 


327 


349 


371 


393 


T 7 
I 8 


224 


244 


265 


285 


306 


326 


346 


366 


2 


210 


229 


248 


267 


287 


306 


324 


344 


a* 


198 


216 


234 


252 


270 


288 


306 


323 


al 


187 


204 


221 


238 


255 


272 


289 


306 


at 


177 


193 


210 


225 


241 


257 


273 


290 


a| 


168 


183 


199 


214 


229 


244 


260 


275 


at 


160 


175 


189 


204 


218 


233 


248 


262 


at 


153 


167 


181 


194 


208 


222 


236 


250 


a| 


146 


159 


173 


186 


199 


213 


226 


239 


3 J 


140 


153 


166 


178 


191 


204 


216 


229 



248 



INDUSTRIAL MATHEMATICS 



TABLE X 
Weights and Areas of Round, Square and Hexagon Steel 

Weight of one cubic inch = .2836 lbs. 
Weight of one cubic toot = 490 lbs. 





Area = 


= Diam. 2 s 


•7854 


Area = 


Side 2 x 1 


Area = 
2 x , 


Diam. 
366 


Thickness 
or 


Round 


Square 


Hexagon 


Diameter 


















Weight - 


Area 


Circum- 


Weight 


Area 


Weight 


Area 




Per 


Square 


ference 


Per 


Square 


Per 


Square 




Inch 


Inches 


Inches 


Inch 


Inches 


Inch 


Inches 


1 

32 


.0002 


.0008 


.0981 


.0003 


.0010 


.0002 


.0008 


& 


.0009 


.0031 


.1963 


.0011 


•0039 


.0010 


.0034 


■h 


.0020 


.0069 


.2995 


.0025 


.0088 


.0022 


.0076 


1 
8 


.0035 


.0123 


•3927 


.0044 


.0156 


.OO38 


•0135 


& 


.0054 


.0192 


.4908 


.0069 


.0244 


.0060 


.0211 


A 


.0078 


.0276 


.5890 


.0101 


•0352 


.0086 


.0304 


£ 


.0107 


.0376 


.6872 


.0136 


.0479 


.OIl8 


.0414 


1 
4 


J)I39 


.0491 


•7854 


.0177 


.0625 


.0154 


.0540 


9 

32 


.0176 


.0621 


•8835 


.0224 


.0791 


.OI94 


.0686 


_5 
16 


.0218 


.0767 


.9817 


.0277 


•0977 


.0240 


.O846 


11 
32 


.0263 


.0928 


I.0799 


.0335 


.1182 


.0290 


.1023 


3 

8 


.0313 


.1104 


I.1781 


.0405 


.1406 


•0345 


.1218 


JL3 
32 


.0368 


.1296 


I.2762 


.0466 


.1651 


.0405 


.1428 


7 
16 


.0426 


.1503 


1-3744 


.0543 


.1914 


.0470 


.1658 


15 
32 


.0489 


.1726 


I.4726 


.0623 


.2197 


.0540 


.1903 


1 
2 


•0557 


.1963 


1.5708 


.0709 


.2500 


.0614 


• 2l6l 


17 
32 


.0629 


.2217 


I.6689 


.0800 


.2822 


.0693 


•2444 


9 
16 


.0705 


.2485 


1. 7671 


.0897 


.3164 


.0777 


•2 743 


19 
32 


.0785 


.2769 


I.8653 


.1036 


.3526 


.0866 


•3053 


5 

8 


.0870 


.3068 


1.9635 


.1108 


.3906 


•0959 


.3383 


21 
T2 


•0959 


.3382 


2.0616 


.1221 


.4307 


.1058 


•3730 


H 


• 1053 


•3712 


2.1598 


.1340 


.4727 


.1161 


•4093 


23. 
32 


.1151 


■4057 


2.2580 


.1465 


.5166 


.1270 


•4474 


3 

4 


.1253 


.4418 


2.3562 


.1622 


.5625 


.1382 


.4871 


25 
32 


•1359 


•4794 


2-4543 


.1732 


.6103 


.1499 


.5286 


13 
T6 


.1470 


.5185 


2.5525 


.1872 


.6602 


.1620 


•5712 


27 
32 


.1586 


•5591 


2.6507 


.2019 


.7119 


.1749 


.6165 


7 
8 


•1705 


.6013 


2.7489 


.2171 


.7656 


.1880 


.6631 


29 
32 


.1829 


.6450 


2.8470 


.2329 


.8213 


.2015 


.7112 


15 
16 


.1958 


.6903 


2.9452 


.2492 


.8789 


.2159 


.7612 


31 
32 


.2090 


•7371 


3-0434 


.2661 


.9384 


•2305 


.8127 


I 


.2227 


.7854 


3.I4I6 


.2683 


1. 0000 


.2456 


.8643 



WEIGHTS AND AREAS 



249 



TABLE X — (Continued) 





Area 


= Diam. 2 x 7854 


Area = 


Side 2 x 1 


Area = 
2 x . 


Diam. 
866 


Thickness 
or 


Round 


Square 


Hexagon 


Diameter 


Weight 
Per 
Inch 


Area 
Square 
Inches 


Circum- 
ference 
Inches 


Weight 
Per 
Inch 


Area 
Square 
Inches 


Weight 
Per 
Inch 


Area 
Square 
Inches 


if 
l& 


.2515 
.2819 

.3141 
.348o 


.8866 

.9940 

1. 1075 

1.2272 


3-3379 
3-5343 
3-7306 
3.9270 


.3201 

.3589 
.4142 

•4431 


I. 1289 
I.2656 
1.4102 
.15625 


•2773 
•3109 
.3464 
.3838 


.9776 
1-0973 
1. 2212 
I-353I 


if 
iA 


.3837 
.4211 
.4603 
.5012 


1.3530 

1.4849 
.16230 
1. 7671 


4-1233 
4-3197 
4.5160 
4.7124 


.4885 
.5362 
.5860 
.6487 


I.7227 
1.8906 
2.0664 
2.2500 


.4231 

•4643 
.5076 
.5526 


I-49I9 
1.6373 
1.7898 
1.9485 


is 

If 


•5438 
.5882 

.6343 
.6821 


I-9I75 
2.0739 
2.2365 
2.4053 


4.9087 
5.1051 
5-3014 
5-4978 


.6930 

.7489 
.8076 
.8685 


2.44M 

2.6406 
2.8477 
3.0625 


.5996 
.6480 
.6994 
•7521 


2.1143 

2.2847 
2.4662 
2.6522 


il 

IT6 
2 


.7317 
.7831 
.8361 
.8910 


2.5802 
2.7612 
2.9483 
3.1416 


5.6941 
5-8905 
6.0868 
6.2832 


.9316 

.9970 

I.0646 

1. 1342 


3.2852 
3.5156 
3-7539 
4.0000 


.8069 

.8635 
.9220 
.9825 


2.8450 
3-0446 
3.2509 
3-4573 


2^ 
2j 


•9475 
1.0058 
1.0658 
1. 1276 


3-3410 
3-5466 
3-7583 
3-976i 


6.4795 
6.6759 
6.8722 
7.0686 


I.2064 
I.2806 
I.3570 
1-4357 


4-2539 
4-5I56 
4.7852 
5.0625 


I.0448 
I.IO91 

I-I753 
1.2434 


3.6840 
3.9106 
4.1440 
4-3892 


2^ 

2| 

2^ 

2i 


1.1911 
1.2564 
1-3234 
1.3921 


4.2000 
4.4301 
4.6664 
4.9087 


7.2649 
7.46I3 
7-6575 
7.8540 


1-5165 
1.6569 
1.6849 
1.7724 


5-3477 
5.6406 
5-9414 
6.2500 


I-3I35 

I.3854 
1.4593 
I-535I 


4-6312 
4.8849 
5-1454 
5-4126 


2f 
2| 
2| 

3 


1-5348 
1.6845 
1.8411 
2.0046 


5-4II9 
5.9396 
6.4918 
7.0686 


8.2467 
8.6394 
9-0321 
9.4248 


I-954I 
2.1446 
2.3441 
2.5548 


6.8906 
7-5625 
8.2656 
9.0000 


1.6924 

1.8574 
2.0304 
2.2105 


5-9674 
6-5493 
7-1590 
7-7941 


3i 
3i 
3l 
3l 


2.1752 
2.3527 
2.5371 
2.7286 


7.6699 
8.2958 
8.9462 
9.6211 


9.8175 
10.2102 
10.6029 
10.9956 


2.7719 
2.9954 
3-2303 
3-4740 


9.7656 
10.5625 
11.3906 
12.2500 


2.3986 
2.5918 
2.7977 
3-0083 


8-4573 

9-1387 

9.8646 

10.6089 



18 



250 



INDUSTRIAL MATHEMATICS 



TABLF X— (Continued) 





Area = Diam. 2 x .7854 


Area = Side 2 x 1 


Area = Diam. 
2 x .866 


Thickness 
or 


Round 


Square 


Hexagon 


Diameter 


Weight 
Per 
Inch 


Area 
Square 
Inches 


Circum- 
ference 
Inches 


Weight 
Per 
Inch 


Area 
Square 
Inches 


Weight 
Per 
Inch 


Area 
Square 
Inches 


3s 
3f 

si 
4 

,i 

4s 

4* 
4f 

4i 

4f 

4l 
4* 

5 

si- 
si 

5f 

si 

58 
-3 
54 

si 

6 

6i 
6i 

6f 

7 

7^ 
8 


2.9269 
3.1323 
3-3446 
3.5638 

3.79OO 
4-0232 
4-2634 
4-5105 

4-7645 
5-0255 
5.2935 
5.5685 

5-8504 
6.1392 
6-4351 
6.7379 

7.0476 

7.3643 
7.6880 
8.0186 

8.7007 

9*4107 

10.1485 

10.9142 

12.5291 
14-2553 


10.3206 
11.0447 
11.7932 
12.5664 

13.3640 
14.1863 
I5.0332 
I5.9043 

16.8002 
17.7205 
18.6655 
19.6350 

20.6290 
21.6475 
22.6905 
23.7583 

24.8505 
25.9672 
27.1085 
28.2743 

30.6796 
33.I83I 
35.7847 
38.4845 

44.1786 
50.2655 


H.3883 
II. 7810 
12.1737 
12.5664 

12.9591 
I3.35I8 
13-7445 
14.1372 

14.5299 
14.9226 

I5.3I53 
15.7080 

16.1007 

16.4934 
16.8861 
17.2788 

17.6715 
18.0642 
18.4569 
18.8496 

19.6350 
20.4204 
21.2058 
21.9912 

23.5620 
25.1328 


3.7265 
3.9880 
4.2582 
4-5374 

4.8254 
5.1223 
5.4280 
5.7426 

6.0662 
6.6276 
6.7397 
7.0897 

7.4496 
7.8164 
8.1930 
8.5786 

8.9729 

9.3762 

9.7883 

IO.2192 

11.0877 
II. 9817 
12.9211 
13.8960 

15.9520 
I8.I497 


13.1407 
14.0625 
15.0156 
16.0000 

17.0156 
18.0625 
19.1406 
20.2500 

21.3906 
22.5625 
23.7656 
25.OOOO 

26.2656 
27.5624 
28.8906 
30.2500 

31.6406 
33-0625 
34.5156 
36.OOOO 

39.0625 
42.2500 
45-5625 
49.OOOO 

56.2500 
64.OOOO 


3.2275 
3-4539 
3.6880 
3-9298 

4.1792 
4.4364 
4.7011 
4.9736 

5.2538 
5-5416 
5-8371 
6.1403 

6.45H 
6.7697 

7-0959 
7.4298 

7.7713 
8.1214 

8.4774 
8.8420 

9-5943 
10.3673 
11. 1908 
12.0351 

13.8158 
15.7192 


11.3798 
12.1785 
13-0035 
13.8292 

14.7359 
15.6424 
16.5761 
17.5569 

18.5249 
19-5397 
20.5816 
21.6503 

22.7456 
23.8696 
25.0198 
26.1971 

27.4013 
28.6361 
29.8913 
3I-I765 

33-8291 
36.5547 
39-4584 
42.4354 

48.7142 
SS.3I69 



Multiply above weights by .993 for wrought iron, .918 for cast iron 
1. 0331 for cast brass, 1.1209 for copper, 1.1748 for phos. bronze, and 
,3265 for aluminum. 



STANDARD DIMENSIONS OF IRON TUBES 



251 



TABLE XI 





Circumference and Area of Circles (1/16" to 2' 


') 


Diam. 


Circum. 


Area 


Diam. 


Circum. 


Area 


A 


O.19635 


0.00307 


I* 


3-3379 


0.8866 


8 


O.39270 


0.01227 


if 


3-5343 


0.9940 


A 


O.58905 


0.02761 


I* 


3-7306 


I.IO75 


i 


O.78540 


0.04909 


I* 


3.9270 


1.2272 


A 


0.98175 


O.07670 


lA 


4.1233 


1-3530 


8 


I.1781 


0.11045 


I# 


4-3197 


1.4849 


7 
16 


1-3744 


0.15033 


iA 


4.5160 


1.6230 


t 


1.5708 


O.19635 


1* 


4.7124 


1.7671 


A 


1. 7671 


0.24850 


** 


4.9087 


I-9I75 


8 


I-9635 


0.30680 


i* 


5-I05I 


2-0739 


11 
16 


2.1598 


0.37122 


i*t 


5-3014 


2.2365 


4 


2.3562 


O.44179 


if 


5-4978 


2.4053 


13 
16 


2.5525 


O.51849 


itt 


5.6941 


2.5802 


8 


2.7489 


O.60132 


1* 


5.8905 


2.7612 


15 
16 


2.9452 


O.69029 


I 16 


6.0868 


2.9483 


I 


3-I4I6 


O.7854 


2 


6.2832 


3.1416 



TABLE XII 

Standard Dimensions of Wrought Iron Welded Tubes. 

Briggs Standard 



Diameter of Tube 




Threaded End 












Nominal 
Inside 


Actual 

Inside 


Actual 
Outside 


Metal 


Number of 
Threads 
per Inch 


Length of 
Perfect 
Thread 


1 in. 


0.270 in. 


0.405 in. 


0.068 in. 


27 


0.19 in. 


I in. 


0.364 in. 


0.540 in. 


0.088 in. 


18 


0.29 in. 


fin. 


0.494 ni. 


0.675 in. 


0.091 in. 


18 


0.30 in. 


1 in. 


0.623 in. 


0.840 in. 


0.109 in. 


14 


0.39 in. 


fin. 


0.824 in. 


1.050 in. 


0.113 in. 


14 


0.40 in. 


1 in. 


1.048 in. 


1.315 in. 


0.134 m. 


II* 


0.51 in. 


i\ in. 


1.380 in. 


1.660 in. 


0.140 in. 


n| 


0.54 in. 


i| in. 


1. 610 in. 


1.900 in. 


0.145 m. 


III 


0.55 m. 


2 in. 


2.067 in. 


2.375 m. 


0.154 in. 


III 


0.58 in. 


2! in. 


2.468 in. 


2.875 in. 


0.204 m. 


8 


0.89 in. 


3 m. 


3.067 in. 


3-500 in. 


0.217 in. 


8 


0.95 in. 


3§ in. 


3-548 in. 


4.000 in. 


0.226 in. 


8 


1.00 in. 


4 m. 


4.026 in. 


4.500 in. 


0.237 in. 


8 


1.05 in. 


42 in. 


4.508 in. 


5.000 in. 


0.246 in. 


8 


1. 10 in. 


5 in. 


5.045 in. 


5.563 in. 


0.259 m. 


8 


1. 16 in. 


6 in. 


6.065 in- 


6.625 in. 


0.280 in. 


8 


1.26 in. 


7 in. 


7.023 in. 


7.625 in. 


0.301 in. 


8 


1.36 in. 


8 in. 


7.982 in. 


8.625 in. 


0.322 in. 


8 


1.46 in. 


9 in. 


9.000 in. 


9.688 in. 


0.344 m. 


8 


1-57 m. 


10 m. 


10.019 in. 


10.750 in. 


0.366 in. 


8 


1.68 in. 



252 



INDUSTRIAL MATHEMATICS 



TABLE XII— (Continued) 

The Sizes of Twist Drills to Be Used in Drilling Holes to Be Reamed 
with Pipe Reamer, and Threaded with Pipe Tap, are as Follows : 



Size Tap 


Diameter Drill 


Size Tap 


Diameter Drill 


i inch 


flinch 


I inch 


ii^ inch 


f inch 


flinch 


\\ inch 


iff inch 


f inch 


Minch 


\\ inch 


iff inch 


\ inch 


ff inch 


2 inch 


2^ inch 


f inch 


M inch 


2§ inch 


2^ inch 






3 inch 


3TS inch 



The Word Diameter, when Used for Gas or Steam Pipe Measure, 
Always Means Inside Diameter 



TABLE XIII 
Sizes of Tap Drills for Taps with U. S. Standard Thread 



Size of 


No. of 
Threads 


Size of 


Size of 


No. of 
Threads 


Size of 


Tap 


per Inch 


Drill 


Tap 


per Inch 


Drill 


i 

4 


20 


A 


li- 


7 


I* 


A 


18 


C 


lt 


6 


iH 


3 

8 


16 


N 


I* 


6 


X 64 


7 
16 


14 


S 


if 


5§ 


Iff 


1 
2 


13 


13. 
32 


if 


5 


ii 


A 


12 


29 
64 


I* 


5 


if 


5 
8 


II 


33 
64 


2 


4l 


T 23 
1 32 


11 

16 


II 


« 


a* 


4* 


-.27 
132 


1 


10 


5 
8 


ai 


4^ 


x 31 
I 32 


H 


IO 


11 
16 


af 


4 


23^ 


1 


9 


47 
64 


H 


4 


2^ 


15 

16 


9 


fi 








I 


8 


27 
32 








If 


7 


61 
64 









TWIST DRILL AND STEEL WIRE GAGES 



253 



TABLE XIV 

Decimal Equivalents of the Number of Twist Drills and 

Steel Wire Gage 





Size of No. 




Size of No. 




Size of No. 




Size of No. 


No. 


in Decimals 


No. 


in Decimals 


No. 


in Decimals 


No. 


in Decimals 


1 


.2280 


21 


.1590 


41 


.0960 


61 


.0390 


2 


.2210 


22 


•I570 


42 


.0935 


62 


.0380 


3 


.2130 


23 


.1540 


43 


.0890 


63 


.O370 


4 


.2090 


24 


.1520 


44 


.0860 


64 


.0360 


5 


.2055 


25 


•1495 


45 


.0820 


65 


.0350 


6 


.2040 


26 


.1470 


46 


.0810 


66 


.0330 


7 


.2010 


27 


.1440 


47 


.0785 


67 


.0320 


8 


.1990 


28 


.1405 


48 


.0760 


68 


.0310 


9 


.i960 


29 


.1360 


49 


.0730 


69 


.02925 


10 


• 1935 


30 


.1285 


50 


.0700 


70 


.0280 


11 


.1910 


31 


.1200 


51 


.0670 


71 


.0260 


12 


.1890 


32 


.1160 


52 


.0635 


72 


.0250 


13 


.1850 


33 


.1130 


53 


•0595 


73 


.0240 


14 


.1820 


34 


.1110 


54 


•055O 


74 


.0225 


IS 


.1800 


35 


.1100 


55 


.0520 


75 


.02IO 


16 


.1770 


36 


.1065 


56 


.0465 


76 


.0200 


17 


.1730 


37 


.1040 


57 


.0430 


77 


.0180 


18 


.1695 


38 


.1015 


58 


.0420 


78 


.0160 


19 


.1660 


39 


.0995 


59 


.0410 


79 


.0145 


20 


.1610 


40 


.0980 


60 


.0400 


80 


•0135 



Letter Sizes 




A 234 

B 238 

C 242 

D 246 

E 25 

F 257 

G 261 



U 368 

V 377 

W 386 

X 397 

Y 404 

Z 413 



254 



INDUSTRIAL MATHEMATICS 




TABLE XV 
Standard Key Seats 



Diameter 


O 


P 


Diameter 


O 


P 


of Hole, 


Width of 


Depth of 


of Hole, 


Width of 


Depth of 


Inches 


Key Seat 


Key Seat 


Inches 


Key Seat 


Key Seat 


3 

8 


A 


3 

64 


3t to 3t£ 


7 
8 


A 


A to | 


i 

8 


A 


3f to 4A 


I 


i 

2 


A to f 


¥ 


5 


4¥ to 4T6 


if 


A 


if to f 


A 


A 


4f to sA 


li 


f 


if to I 


A 


7 
64 


5i to Sif 


if 


ii 

16 


ff toif 


i 

4 


1 


5l to 6& 


If 


4 


Ills to if 


5 
16 


A 


61 to 6ff 


If 


13 
16 


iA to if 


8 


A 


6f to 7A 


If 


7 


iii to if 


A 


A 


71 to 7 H 


If 


15 
16 


iff to 2f 


I 


i 


7f to 8^ 


2 


I 


2^ tO 2if 


f 


S 

16 


8f to 8ff 


2f 


iA 


2f to 3A 


4 


f 


9 to io 


2| 


if 



MULTIPLICATION TABLES 



: DD 



TABLE XVI 
Multiplication Tables 25 X 20 



= =?00 



/ 


2 


3 


4 


5 


6 


7 


8 


9 


10 


II 


12 


13 


14 


IS 


16 


17 


18 


19 


20 


2 


4 


6 


8 


10 


12 


14 


16 


18 


20 


22 


24 


26 


28 


30 


32 


34 


36_ 


38 


40 


3 


6 


9 


12 


15 


18 


21 


24 


27 


30 


33 


36 


39 


42 


45 





51 


54 


57 


60 


4 


8 


12 


16 


20 


24 


28 


32 


36 


40 


44 


48/ 


"52 


56 


60 


64 


66 


72 


76 


80 


5 


10 


15 


20 


25 


30 


35 


40 


45 


SOj 


55 


60 


65 


70 


75 


80 


85 


90 


£L 


IOO_ 


6 


12 


18 


24 


30 


36 


42 


48/ 


'U 


60 


66 


72 


78 


84 


90 


96s 


102 


108 


1/4 


120 


7 


14 


21 


28 


35 


42 


49 r' 


'56 


63 


70 


77 


84 


91 


98,' 


105 


112 


1/9 


726 


133 


140 


8 


16 


24 


32 


40 


4ej 


56' 


64 


72 


80 


88 


96/ 


f l04 


112 


120 


726 


136 


144 


152 


160 


9 


18 


27 


36 


45 r 


54 


63 


72 


81 


90 


9? 


108 


117 


126 


135 


144 


153 


162 


171 


180 


10 


20 


30 


40 


5$ 


60 


70 


80 


90 


I0Q,' 


I/O 


120 


130 


140 


150 


160 


170 


180 


190 


200_ 


II 


22 


33 


44 1 


'55 


66 


77 


68 


gg. 


'no 


121 


132 


143 


IS4 


165 


176 


187 


19/ 


209 


220 


12 


24 


36 


481 


60 


72 


84 


9f 


108 


120 


132 


744 


IS6 


168 


ISO 


192) 


204 


2/6 


226 


240 


IZ 


26 


M 


52 


65 


78 


»t 


104 


117 


130 


143 


156 


169 


182 


/9S> 


'208 


221 


234 


247 


260 


14 


26 


4?\ 
1 


56 


70 


84 


981 


112 


126 


140 


154 


166 


182 


I9& 


'210 


724 


238 


252 


266 


280 


IS 


30 


*i 


60 


75 


90 1 


10S 


120 


135 


ISO 


I6S 


180 


199 


'2/0 


225 


240 


255 


270 


285 


■300 


16 


32 


4 


64 


80 


96} 


1/2 


126 


144 


160 


176 


19? 


208 


224 


240 


256 


272 


288* 


304 


320 


17 


34. 


SI 


68 


651 


'702 


1/9 


/36 


153 


770 


/67( 


204 


22/ 


238 


255 


272 


289 


'306 


323 


340 


18 


36\ 


54 


72 


90\ 


108 


126 


144 


162 


180 


&4 


2/6 


234 


252 


270 


268/ 


306 


324 


342 


360 


19 


36 i 


57 


76 


95 \ 


114 


133 


152 


I7J 


I90> 


"209 


228 


247 


266 


285, 


304 


323 


342 


361 


380 


20 


40 \ 


60 


80 


100} 


120 


140 


160 


180 


200) 


220 


240 


260 


280 


300} 


320 


340 


360 


380 


400 


21 


«l 


63 


84 i 


'tos 


126 


147 


763 


169 1 


f 2/0 


231 


252 


273 


294. 


<S,5 


336 


357 


378 


399 


'420 


22 


1 
44\ 

j 


66 


8d\ 


110 


132 


154 


176 


19$ 


220 


242 


264 


286{ 


308 


330 


352 


374 


396 


4/8 


440 


23 


«| 


69 


92\ 


115 


138 


167 


784\ 


207 


230 


253 


276 


299) 


322 


345 


368 


39/ 


U/4 


437 


460 


24 


«| 


72 


*\ 


120 


144 


168 


I92\ 


2/6 


240 


264 


286* 


3/2 


336 


360 


3641 


'408 


432 


4S6 


480 


25 


50 j 


75 


IOC\ 


125 


150 


175 


20<\ 


225 


2S0 


275 


30a 


32S 


350 


375 


40d 


425 


450 


47S 


SOO 



50 



100 



= = 300 



400 



50 



100 



200 



300 



400 



BOO 



256 



INDUSTRIAL MATHEMATICS 




Cylinder Cover 

Cylinder 

Main Valve Rocker 

Main Valve Stem Gland- 
Valve Chamber Cover- 

Main Valve and Bushing — 
inside of Chamber 

Piston Gland Stud 

Piston Gland 

Bumper Spring-- •"''. 

Bumper Cap- — " 
Piston Rod--— 
Guide Shoe. L. ft.- 

Guide Stud Washer^- - 

Incline dovetailed N vJ. 
in back of Ram 
Ram Guide. L./i. 
Safety Pin 
Ram 

Ram Die Key-~~'i~. 

Ram Die— 

Anvil Die 

Anvil Cap— 



■Main Valve Link 
-Main Valve Couplinc/ 
Main Valve Stem 
■Throttle Yoke 
■Throttle Stem Gland 

■ Throttle Valve and Spring 
inside Chamber 

Steam Pipe Nipple & Gland 
■Throttle Stem in back 

-Exhaust Pipe Nipple 

Exhaust Gland 

Throttle Lever 
• Cam & Cam Lever inside of Frame 

■ Cam Arm & Main Valve 
Connection back of Frame 

•Guide Shoe. R. fi. 
Operating Lever 

Segment 
\.Lever Latch 



■Lag Screw 



Tree Nail 



Steam Hammer 




Planer 



A— Bed. 

B — Reversing lever. 

C — Feed friction. 

D — Belt shifting mechanism. 

E — Hand cross feed lever. 



F — Vertical feed levers. 
G — Tool head. 
H — Cross rail. 
I — Table or platen. 
J — Reverse dog. 




Boring Mill 



ANSWERS 

ANSWERS TO MATHEMATICS 

Notation and Numeration — Page 3 
5- 85. 

8. i, 648, 432. 

Addition, Subtraction, Multiplication and Division — Page 8 

1. (a) 880,369. (6) 4,866. 

2. (a) 32,983. (&) 5.997- 

3. (a) 774- (&) 60,727. 

4. (a) 86,081. (b) 342. 
5- (a) 35- 0) 339- 
6. (a) 9,990. (6) 1,259. 
7- (a) 7.056. (6) 2,091. 

8. (a) 22,960. (6) 98,000. 

9. (a) 1,566,585. (ft) 8,104,320. 

10. (a) 843. (6) 162. 

11. (a) 23. (6) 64,305. 

12. (a) 10,770. (6) 1,009. 

Cancellation and Least Common Multiple — Page 9 



(&) 48. 

(6) 81. 

(b) 90. 

(6) 180. 

Common Fractions — Page 13 
6|; 27I; 2 6/83; 25 1/5; 1 131/763; 1 2/121; 1 1/12. 

2. 21/4; 73/9; 20/3; 103/8; 127/9; 55/8; 143/12; 2449/49; 790320/889; 
2134/17- 

3. 1/6; 9/25; 1/18; 4/15; 5/12; 3/8; 5/8; 40/323; 47069/67598; 32/143. 

4. 161/63; 1 23/352; 1 5/76; 224/225; 9/16; 1/20; 1/4; 14/15; 1 i/3- 

257 



I. 


120. 


2. 


280. 


3- 


24. 


4. 


(a) 252. 


5- 


(a) 60. 


6. 


(a) 72. 


7- 


(a) 60. 


I. 


26f; 3 i/7; 8| 



258 



INDUSTRIAL MATHEMATICS 



28/4; 36/2; 36/3; 96/12; 50/10; 18/3; 32/8; 42/V, 20/5; 221/13. 
6/7; i/3; 31/32; 61/3; 14 1/4; 18 1/2; 12430/37; 5I/I44; 14 3/7. 
40/64; 32/48; 12/128; 60/64; 65/160; 12/8; 200/24; 39/6. 
12/24, 8/24, 9/24; 160/180, 135/180, 72/180; 8 1/8, 3 5/8, 7/8; 
1/16, 516/48; 32/36, 9/36, 218/36; 52/16, 34/16, 1/16; 24/64, 
5 1/64, 28/64; 8 8/32, 2 16/32, 9/32; -832/64, 9/64- 



(a) 8 19/24. 
(a) 865 11/24. 
(a) 88 389/420. 
(a) 7 3/8. 
(a) 6 17/24- 
(a) 25 37/39- 
(a) 4/15- 
(a) 1/4. 
(a) 14 21/128. 
(a) 1 1/6. 
(a) 1/2. 
(a) 60 3/5- 
8 23/32 ins. 

2 43/64 ins. 
61/4 hrs. 
3/8 ft. 
$19-35- 

10 5/8 ft. 

4/45- 

3 21/22. 

145- 
9- 

(a) 51,497.50. 

(a) 126.375. 

(a) 2,749.3709467 

(a) 79-375- 

(o) 0.539991. 

(a) 0.611. 

(a) 124.74. 

(a) 697-5- 

(a) 0.46875. 



(6) 48 13/24- 

(&) 82 135/182. 

(b) 512 37/48. 

00 5/6- 

(6) 61/64. 

(6) 9761 7/8. 

(b) 2 11/32. 

(6) 182 13951/16384. 

(6) 4670 13/40. 

(*) 2. 

(6) 4 127/158. 

(b) 5445/1868. 



Decimal Fractions — Page 18 

(b) 86,086.3. 

(6) 46,494-38793- 

(6) 1064.62638. 

.(b) 0.13503. 

(&) 0.058483. 

(&) 273.397. 

(ft) 13,777-44. 

(b) 0.6889. 

(ft) 133.438. 



ANSWERS TO EXERCISES 



259 



10. 


(a) 1.820. 






(b) 


10.762. 


II. 


(a) 1,000. 






(b) 


140,005,000. 


12. 


(a) 0.0002. 






(b) 


1.605. 


13- 


(a) 0.0156; 0. 


125. 




(b) 


0.7143; 0.375- 


14. 


(a) 0.1094; 0. 


8889. 




(b) 


0.6667; 4-8333- 


15- 


(a) 1/2; 3/4. 






(b) 


2 5/8; 4 13/32. 


16. 


(0) 3 35/64; 7 3/i6. 




(b) 


25 3/64; 1 3/8. 


17. 


$7,525. 










18. 


$3.75. 










19. 


$0,325. 










20. 


8.3125 lbs. 
















Percentage — Page 19 


1. 


$24.00 






6. 


100%. 


2. 


90. 






7- 


263. 


3- 


1.5. 






8. 


370.37. 


4- 


7 159/163%. 






9- 


$285.71. 


5- 


150%. 






10. 


16.8 h.p. 






Weights and 


Measures — Page 22 


1. 


2520 in. 






22. 


6 lbs. 3 oz. 


2. 


5 yds., 1 ft., 8 


in. 




23- 


60 5/12 oz. 


3- 


3 miles. 1 160 ft. 




24. 


1,152,000 gr. 


4- 


11 miles, 4620 


ft. 




25- 


10 ser. 


5- 


52 ft. 6 in. 






26. 


25 dr. 


6. 


3840 acres. 






27. 


7 lbs., 5 7/12 oz. 


7- 


94 sq. in. 






28. 


84 oz. 


8. 


26,666 sq. yds 


. 6 sq. 


ft. 


29. 


792 dr. 


9- 


11,025 sq. ft. 






30. 


8 bbls. 


10. 


4,200 sq. rds. 






31. 


21 qts. 


11. 


52.92 cu. ft. 






32. 


2,016 gills. 


12. 


221,184 cu. in. 






33- 


1 bu. 


13- 


2,376 cu. ft. 






34. 


4096 pints. 


14. 


250,560 cu. in. 






35- 


5 gal. 20 gills. 


15. 


113,280 gr. 






36. 


144 qts. 


16. 


32,000 oz. 






37- 


100 pecks. 


17- 


93 tons, 24,000 oz. 




38. 


2560 pints. 


18. 


100 cwts. 






39- 


324,000 sec. 


19. 


102,768 oz. 






40. 


6 deg. 40 min. 


20. 


i6f pwt. 






41. 


290,700 sec. 


21. 


104 oz. 80 gr. 






42. 


60 min. 



260 INDUSTRIAL MATHEMATICS 



43. 1,296,000 sec 




47. 648 hrs. 38,800 min. 


44- 5,ii3l days. 




48. 1,051,920 min. 


45. 9 yrs. 




49. 104.8 m.m. 


46. 2,700 min. 




50. 4-7244". 




Ratio and Proportion — Page 25 


1. 533i 




9. 2\ days. 


2. 2.36 gal. 




10. 902.87 days. 


3- $i,333i 




11. 90 men. 


4. $.45. 




12. $319-3- 


5- 24 T. 




13. $0,196. 


6. 10 days. 




14. 3 3/224 days. 


7. 16 days. 




15. 99.5 oz. 


8. 13^ days. 








Taper Calculations — Page 27 


1. 0.052". 




6. 0.03125" or 1/32". 


2. 0.150". 




7. 0.125". 


3. 0.050". 




8. 0.350". 


4. 0.130". 




9- 0.139". 


5. 0.250". 




10. 0.078". 
Interest — Page 29 


I. $70.83. 




6. $232.81. 


2. $276.45. 




7. $5I7.44. 


3. $603.75. 




8. $571.20. 


4. $57i-io. 




9. $3,540.87. 


5. $1,371.60. 




10. $1,238.99. 


] 


Pulley and Gear Diameters — Page 31 


1. 256 r.p.m. 




7. 1000; 480; 257^; 13 


2. 616 r.p.m. 




30; 16 1/14; 8£. 


3. i66| r.p.m. 




8. 729 to 1. 


4- 32 to 1. 




9. 13.78; 22.96; 38.27 


5. 20 to 9. 




108; 180; 300; 500. 


6. 6 in. 




10. 83^ and 22.7 f.p.m. 


Square Root, 


Involution and Evolution — Page 35 


1. 496. 




6. 1,111. 


2. 364. 




7. 387,420,489. 


3. 222. 




8. 6. 


A- 157. 




9- 823,543. 


5. 21.5139. 




10. 15.625. 



1331; 62!; 



63.78; 



ANSWERS TO EXERCISES 



26l 







Square 


Root and Triangulation — Page 36 


I. 


16.155 ft. 




8. 5.887 in. 


2. 


121.037 ft. 




9. 3.250 in. 


3- 


224.4 ft. 




10. 3.606 in. 


4- 


0.53 in. 




11. 9.540 cu. in. 


5- 


180.278 ft. 




12. 5-385 ft. 


6. 


2.862 in. 




13. 5 ft- 


7- 


8-347 in. 




14. 77-518 in. 



I- 33.076,161. 

2. 1,953,125. 

3- 161,051. 

4. 29. 



Cube Root — Page 39 

5- 999- 
6. 46.369. 
7- 0.75- 



I. 


0.663 sq. in. 


2. 


i£ in. 


3- 


42 sec. 


4- 


37.699 in. 


5- 


5.67 pints. 


6. 


282.1 gals. 


7- 


5 in. 



Circles — Page 42 



I. 


9 sq. in. 






13 


395.842 sq. in.; 


518.250 cu. in 


2. 


4,800 sq. ft. 






14 


18.617 cu. yds. 




3- 


$48.00. 






15 


44.108 lbs. 




4. 


$21.75. 






16 


$1,885. 




5- 


12 sq. in.; 7.688 


sq 


. in.; 15 sq. in. 


17 


II3.097 sq. in.; 


43-756 in. 


6. 


14.697 sq. in. 






18 


43.982 cu. in. 




7- 


20 deg.; 72 deg 


.; 


138 deg. 


19 


60 cu. in. 




S. 


28 deg. 






20 


33.6 cu. in. 




9- 


423.014 gal. 






21 


1.302 in. 




10. 


1512. 






22 


2.351 in. 




11. 


1.299 cu. in. 






23 


9.184 in. 




12. 


28.274 sq. in.; 


14.137 cu. in. 


24 


1.663 in. 





262 



INDUSTRIAL MATHEMATICS 



Review Exercises — Page 51 



1. 1,795.20 gal. 

2. 1,584 paces. 

3. 7,920 ft. 

4. $16.00. 

5. 21,6565 lbs. 

6. 385.9 bu. 

7. 1.6 h.p. 

8. io6f ft. 



9. 3.543 in. 

10. 4,500 lbs. 

11. 5.148 gr. 

12. 4,021.45 h.p. 

13. 22.88 lbs. 

14. 113,588,000 ft. lbs. 

15. 4.584 lbs. 

16. 1. 816 in. 



Formulas and Algebraical Expressions — Page 57 

1. (a) ga - 2b + x. (b) 3a + 4b + ZC. 

2. (a) 2a 2 — 2b — 20. (b) 8xy 2 z 3 . 

3. (a) a 2 + 10a + 24. (b) x 2 — y 2 . 

4. (a) 2b. (Z?) 4xy. 

8. 6.333". 

9. 128.57 tons. 
10. 13.708 b.h.p. 



5- 


6£. 


6. 


7- 


7- 


— 2 


11. 


5 = 


12. 


P = 



33000 h.p. 

U.W. 

33 000 h.p. 

L.A.N. 



A -f TTCrf. 

13. a = 7 ; 

■wb 
wa.b. — A 



v = 33000 h.p. 
S.W. 
33000 h.p. 



TTd 



d = 



P.A.N. ' 
= 33 000 h.p. 
P.A.iV. 

A -f- 7TC.d. 

7ra 
7ra.&. — A 



W = 

A = 



33000 h.p. 

S.U. 

33000 h.p. 

P.L.N. 



14. O = 2*7 - 2F.X. - 2T.U.Z. + 2M.C/.Z. + 2 y, ^- Z - . 



W.P. ir W.P. e 



W.P- _ T/ p _ A(V+S) 
A v > r ~ W 



16. t 



I2S 



25 

i 2 



17. V = \2g.S. + 



v = <v 2 



2g..S- 



:8. H = \A 2 + B 2 



19. M 



20. H = 



Vz.s V25 

D(N - M) 



A = \H 2 - B 2 

VzTs 



2V 



N 



MT° r 
5H - £>.M 
_ D ; 



g = 


V 2 - 

2S 


V 2 


B = 


<H 2 - 


-A 2 . 


M ; 


S = 


M 2 .N 2 
Z 


M 


5# 


-D.N 



ANSWERS TO EXERCISES 



263 



1. 4. 

2. 13. 
3- 50. 
4. 8. 
5- 2. 



Progression — Page 63 

6. 729. 

7. 4. 

8. 2. 

9. 29.514. 
10. 1. 15. 



Trigonometry — Page 71 

1. 3-8992 in.; 4.6631 in.; 4.0037 in. 

2. 11.230 in.; 14.1138 in.; 12.1240 in. 

3. 9 deg. 28 min.; 18 deg. 26 min.; 11 deg. 19 min. 

4. 7 deg. 17 min. — R.; 2 deg. 43 min. — L. 

5. 4 deg. 33 min. 8. 0.3492 in. 



6. 115.47 ft. 






9- 


0.5773 in. 


7. 1.6643 miles. 






10 


3.4641 in. 


Trigonometry (Continued) — Page 72 


I. 0.6946 in.; 5.4941 in.; 






13- 


14.874 miles. 


3.1187 in. 






14 


3 deg. 48 min. 


2. 8.6933 in.; 20.8321 in.; 






IS- 


A = 75 deg.; a = 5.796 in.; 


24.0488 in. 








b = 1.553 in. 


3. 22 deg. 1 min.; 25 deg. 


4i 


min.; 


16. 


B = 76 deg.; a = 0.212 in.; 


30 deg. 








c = 0.878 in. 


4. 12 balls. 






17. 


A = 29 deg. 56 min.; B — 60 


5. 1.466 in. 








deg. 4 min.; c = 0.935 in. 


6. 52.303 ft. 






18. 


B - 48 deg.; b = 8.918 in.; 


7. 24 deg. 








a = 8.030 in. 


8. 860.24 ft. 






19. 


1.9109" A-B; 2.0169" A-C; 


9. 1.342 in.; 2.684 in. 








1.732" C-D; 2.4397" C-E. 


10. 1" X 3" X 2.732". 






20. 


0.8505" V; I.5I5S" W; 


11. 60 deg. 42 min. 








0.866" X; 2.375" F; 


12. 1,464.1 ft. 








0.8838" Z. 


Feeds and Speeds — Page 76 


1. 28.798 r.p.m. 






10. 


349.8 r.p.m. 


2. 57.268 ft. 






11. 


78.54 f-P-m. 


3. 6.87 min. 






12. 


125.664 f.p.m. 


4. 2.78 min. 






13. 


5 min. 13 sec. 


5. 31.8 r.p.m. 






14. 


1 min. 28 sec. 


6. 94.248 f.p.m. 






IS- 


8 hrs. 20 min. 


7. 53^ min. 






16. 


29 hrs. 24 min. 


8. 3,819.7 r.p.m. 






17. 


98.2 r.p.m. 


9. 382 r.p.m. 






18. 


108.974 f.p.m. 


19 











264 INDUSTRIAL MATHEMATICS 





Cost Calculation — Page 87 


I. $28.50. 


9- $0.51- 


2. $380.01. 


10. $5.91. 


3- 13-12. 


11. $6.76. 


4- $5-77- 


12. $8.66. 


5- $65.20. 


13. $4.96. 


6. $24.45. 


14. $1.60. 


7. $2.40. 


15. $230.90. 


8. $8.48. 


16. $12.51. 




Levers — Page 90 


1. 52.08 lbs.; 52. 


08 lbs. 6. 16.5 in. 


2. 0.352 ft. 


7. 207.85 lbs. 


3. 400 lbs. 


3. 140 lbs. 


4. 108 lbs. 


9. 173I lbs. 


5- 2 ft. 


10. 4712.4 lbs. 




Pulleys — Page 93 


1. 1. 


6. 480 lbs. 


2. 181.82 lbs. 


7. 100 lbs. 


3- 75 lbs. 


8. 2,880 lbs. 


4. 75 lbs. 


9. 29.41 lbs. 


5- 50 lbs. 


10. 7.4% or 6.667 lbs 




Screws — Page 95 


1. 28,274.4 lbs. 


5. 19,792.08 lbs. 


2. 15,079.68 lbs. 


6. \ in. 


3. 9 in. 


7. 5.68 in. 


4- 3S.I85-92 lbs. 


8. 4. 




Inclined Planes — Page 97 


1. 812.27 lbs. 


6. 678.86 lbs. 


2. 888.89 lbs. 


7. 326.35 lbs. 


3. 9,178.25 lbs. 


8. 5%- 


4. 618.41 lbs. 


9- I5%- 


5. 184.89 lbs. 


10. 442%- 




Wedges — Page 98 


1. 20 lbs. 


4. 4.69 lbs. 


2. 20 ins. 


5. 75 lbs. 


3. 4000 lbs. 





ANSWERS TO EXERCISES 



265 



i. 

2. 

3- 

4- 

5- 
6. 
7- 
8. 
9- 

10. 

ii. 

12. 
13- 
14. 

IS- 

16. 
17. 
18. 



I. 

2. 

3- 
4- 
5- 
6. 
7- 
IS- 



6 P. 

0.250". 
0.100". 
0.100". 

4-250". 

5". 

5i". 

4.200". 

32. 

9.868". 

15.708". 

18.850". 

0.0224". 

3.750". 

0.262". 

4.200". 

0.0157". 

5". 



2.309". 

57 deg. 31 min. 

0.770". 

2 deg. 29 min. 

2 deg. 29 min. 

4.094". 

4". 



Spur Gearing — Page 104 
19. N = 50, C.P. = 0.3142 



= 0.1000", Ded 
Wh.d. = 0.2157' 
5.2000'" 

60 
ar \ P.D. = 6.000" 
6.200 

= 52 
21. Gear-! P.D. =6.500 
6.750 



22. 



T = 0.1571". Add. 
0.1000", Clear. = 0.0157", 
Wg.d. = 0.2000", O.D. = 



fN =60 
\ P.D. = 
I O.D. = 

fN =52 
{ P.D. = 
[ O.D. = 



Pinion 



f N = 40 
inion \ P.D. = 4. 
I O.D. = 4. 
f N =30 
\ P.D. = 3. 



000' 
200' 



Pinion -j F.u. = 3-750' 
[ O.D. = 4.000' 





Driver 


Driven 
on J.S. 


Driver 
on J.S. 


Driven 


P.D 

O.D 


2" 
2.2" 


3" 

3.2" 


1.5" 

i*7" 


3-5" 
3-7" 



23. No. 5. No. 3. 

Bevel Gearing — Page 107 



8. 0.2157". 

9. 0.174" 

10. 45 deg. 10 min. 

11. 26 deg. 42 min. 

12. 0.240". 
13- 3-333" 
14. 7.263". 



D.D. 


Add. and 
Ded. 


Clear. 


Wh.D. 


Wg.D. 


T 


T at Small 
End 


Ang. 
Add. 


5" 


0.1 00" 


0.0157" 


O.2157" 


0.200" 


O.1571 


O.1047 


0.050" 


Width 
of Face 


P.C.R. 


Add. and 
Ded. Angle 


Ded. 

Ang. 


Face 
Angle 


O.D. 


Cutting 
Angle 


O.962" 


2.8868" 


i° 59 7 


i° 59' 


28° 1' 


5.100" 58 2' 



16. Gear 



O.D. = 5.188" 

Face Angle = 27 42' 

P.C. Angle = 68° 12' 

17. Gear -j Face Angle = 20 i' 
Cutting Angle = 66° 25' 

18. No. 2 cutter. 



T5 . . /O.D. =3.286" 
Pimon \F a ceAngle=55°46' 

P.C. Angle =2i° 48' 
i<! Face Angle = 66° 25' 

Cutting Angle = 20 i' 



266 



INDUSTRIAL MATHEMATICS 







Worm 


Gearing — Page in 


I, 


1.906". 




6. 0.885". 


2. 


3-076". 




7. 0.054" 


3- 


3.368". 




8. 10 to 1. 


4. 


2.038". 




9. 0.452. 


5- 


6° 28'. 




10. 



Worm 



Worm Wheel 



Lead = 0.4000" 
Add. = 0.0637" 
Wh.D. = 0.1373" 
Tooth Angle =5° 18' 
P.D. = 1.3727" 
R.D. = 1.2254" 
t I. = 0.062" 

C.P. = 0.2000" 
P.D. - 2.0372" 
Th.D. = 2.1645" 
R. of Throat = 0.6227' 
O.D. = 2.3898" 



Spiral Gearing — Page 114 

1. D = 3.535". = 3.735". W = 0.2157". Add. = 0.100". C = 3-535 . 
L = 11. 107". 

2. DA = 7.201", Da = 2.773". NA = 36", Na = 24", OA = 7.401", 
Oa = 2.973", No. of cutter for A = No. 1, No. of cutter for a = 
No. 3. C = 4-987" 

3- = 3-431". 

4. C = 9.429". 

5. D = 4.713". # = 20, N.C.P. = 0.5236", No. of cutter = No. 2, 

c = 4.713". 

6. Z>A = 3.224", Da = 1.274", OA = 3-391", Oa = 1.441", C.P.A, 

= 0.4221", C.P.a. = 0.3337", P.N.A. = 0.2618", P.N.a. = 0.2618". 
No. of cutter A = 2, No. of cutter a = 5, NA = 0.24, aA =51° 40'. 

7. i| to 1. 

Review Exercises — Page 114 



1. i8f ft. 






9. 502.75 in. 






2. 585 ft. 






10. xi = 1.526 in.; 


X2 


= 2. 181 in.; 


3. $2.06. 






X3 = 2.269 in- 






4. 58.81 lbs. 






11. 24. 






5. 11. 018 in. 






12. 242. 






6, 566 balls- 


-525 & 


500. 


13- 1-375- 






7. 3.215 in. 






14. 11.61 lbs. 






8. 9.78 bbls. 






15. 18.38 lbs. 







ANSWERS TO EXERCISES 



267 



1. 4.100 in. 

2. 0.2925 in. 

3. 8.358 in. 



Dovetails — Page 117 

4. 1.957 in. 

5. 2.167 m - 



Screw Threads — Page 121 



1. V = 0.] 


;o8", 


u.s.s. 


= 0.081". 


9. 0.125", 0.750" 


2. U = 0.072", 


u.s.s. 


= 0.054". 


10. Triple th'd. 0.125". 


3. 0.620". 








11. D = 1.375". P = 0.1666", 


4. 0.892". 








d = 0.1443". PD = 1.2307 1 


5. 0.468". 








RD = 1.0864". 


6. 0.031". 








12. 0.584". 


7. 0.267". 








13. 2.811". 


8. 0.0179" 


, 0.0556". 




14. 2.443" 










15. 2° 17'. 






Lathe Change Gears — Page 124 


1. 65. 








6. 60/50. 


2. 69. 








7- 45- 


3- 30. 








8. 6 2/5. 


4. 5.5o". 








9. 25.6. 


5. 120. 








10. sh 



Indexing — Page 128 

(Note. — Problems in indexing can have more than one answer) 
(Below is shown only one answer for each problem) 



3 turns 5 holes in 15 hole circle. 
1 turn 9 holes in 21 hole circle. 

2 holes in 17 hole circle. 

8 holes in 17 hole circle. 

8 holes in 23 hole circle. 
+21/23-11/33. 
+10/21-10/33. 
+23/29-11/33. 
+6/2-6/33. 
+13/39-3/49- 

Spindle 48, worm 24, 2 idlers. 
Spindle 40, worm 56, 2 idlers. 



14. Spindle 48, 1st stud 24, 2d 
stud 40, worm 72, 1 idler. 

15. Spindle 64, 1st stud 24, 2d 
stud 24, worm 72, no idlers. 

16. 1 turn 8 holes in 18 hole 
circle. 

17. 1 turn 7 holes in 27 hole 
circle. 

18. 16 holes in 27 hole circle. 

19. 2 turns 7 holes in 18 hole 
circle. 

20. 23 holes in 27 hole circle. 



268 



INDUSTRIAL MATHEMATICS 



i. Driven 72X64, driver 24X40. 

2. Driven 72 X64, driver 48 X24. 

3. Driven 24X40, driver 64X100 

4. 10 in. 

5. 12! in. 



Spiral Milling — Page 130 

(One answer for each problem) 
6 



Driven 64 X40, driver 32 X 
ioo, 38 deg. 9 min. 

7. 15 deg. 31 min. 

8. Driven 72 X64, driver 24 X32. 

9. Driven 56 X64, driver 28 X40. 
o. Driven 48 X24, driver 24 X32. 



1. 0.286. 

2. 0.188. 

3. 0.179. 

4. 1 deg. 12 min. 

5. 11 deg. 19 min. 



1. (a) 13.073 ohms. 
(&) 50,484.7 ft. 

2. 0.638 ohms. 

3. 2.4 ohms. 

4. 1.1 ohms. 

5. 8.8 ohms. 

6. 137-5 amps. 

7. 4.4 amps. 



Friction — Page 132 

6. 25 deg. 38 min. 

7. 11 deg. 19 min. 

8. 75 lbs. 

9. 87! lbs. 
10. 33 lbs. 



Electricity- 



Page 134 

8. 11 kilowatts. 

9. 2,000 volts. 

10. 1.5 volts. 

11. 70 volts. 

12. 0.02286 amps. 
13-52 cents. 

14. $13.14. 



Horsepower Calculation — Page 138 



1. 510,625 ft. lbs. 

2. 175 lbs. 

3. 0.141 h.p. 

4. 1,042.55 h.p. 

5. 8.06 h.p. 



6. 0.60 h.p. 

7. 5.641 in. 

8. 12.8 h.p. 

9. 14.77 h.p. 

10. 6.4 h.p. (Formula No. 2). 
5.38 h.p. (Formula No. 3). 



Strength and Proportion of Gear Teeth — Page 139 

1. 47.85 h.p. 6. 2.006 in. 

2. 0.623 h.p. 7. 0.897 h.p. 

3. 3 P or 1 in C.P. cutter. 8. 1.25 h.p. 

4. 4.299 in. 9. 9.958 in. 

5. 2 P cutter. 10. 1 § P cutter. 



ANSWERS TO EXERCISES 



269 



Page 141 

7. 707.11 lbs. 

8. 223.61 lbs., 116 deg. 34 min. 
with 1000 lbs. force in N.W. 
direction. 

9- 5.3I5-07 lbs., 131 deg. 11 min. 

with smaller force in N.W. 

direction, 
o. 50 lbs. 



Resolution of Forces 

1. 49,500,000 ft. lbs. per min. 

2. 600 lbs., no direction (R =0). 

3. 24 min., B's direction or west 
Shore. 

4. Tension 2.446 tons, compression 
9.534 tons. 

5. 70.71 lbs., 135 deg. with A in 
N.W. direction. 

6. 22.36 lbs., 26 deg. 34 min. with 
vertical, on opposite side of the 
other component. 



Falling Bodies — Page 144 

1. 160.8 ft. ' 6. 113.42 ft. per sec, 3.53 

2. 3,618 ft., 14.472 ft. 7. 3. 11 sec, 155-47 ft, 

3. 7.89 sec. 8. 1,608 ft., 321.6 ft. 

4. 80.2 ft. per sec 9. 476.92 ft., 159.71 ft. 

5. 38.87 ft., 155-47 it. 10. 188.94 ft. per sec 

Centrifugal Force — Page 145 

1. 3,068.89 lbs. 6. 721.8 ft. per min. 

2. 1,990.06 lbs. 7. 24.25 ft. 

3. 127.87 lbs. 8. 25.082.92 lbs. 

4. 777-13 lbs. 9- 218.57 lbs. 

5. 244.39 lbs. 10. 170.50 lbs. 

Horsepower of Belting — Page 147 

1. 24 h.p. 6. 49.09 h.p. 

2. 5.566 in. 7. 22.5 h.p. 

3. 35 h.p., 60 h.p. 8. 1.6 in. 

4. 6.111 in. 9- 5-5 hi. 

5. 5.893 in. 10. 2.357 in. 



Length of Belting — Page 148 

1. 59.82 ft. 6. 36.9 ft. 

2. 34-i6 ft. 7- 33.76 ft. 

3. 40.03 ft. 8. 35.18 ft. 
4- 45-35 it. 9- 36.16 ft. 
5. 55.22 ft. 10. 74-8i ft. 



27O INDUSTRIAL MATHEMATICS 

Rope Drives— Page 150 

1. 15 h.p. 6. 5 ropes. 

2. 60 h.p. 7. 13 ropes. 

3. 234.38 lbs. 8. 487.5 lbs. 

4. 0.675 lbs., 2.7 lbs. 9. 5 ropes. 
5- 339-29 h.p. 10. 3.288 in. 

Cable or Wire Rope Drives — Page 151 

1. 300 h.p. 6. 216.46 h.p. 

2. 187.5 h.p. 7. 1 cable. 

3. 1.58 lbs., 3.57 lbs. 8. 2 +cables. 

4. 622.125 lbs. 9. 1.558 in. 

5. 1 cable. 10. 26.77 lbs. 

Chain Transmission — Page 153 

1. 3-236". 4- P.D. = 14-349". O.D. = 

2. O.D. = 6.084", B.D. = 5.434" 14.837". B.D. = 13.861". 

3. A = 1.050", B = 0.700", 5. 7.201". 
b = 0.569". 

Shaft Design — Page 156 

1. 1.08 deg. (1 deg. 4.8 min.). 6. 12.66 h.p. 

2. 1. 138 in. 7. 7.65 in. 

3. 70 h.p. 8. 0.0284 deg. (1.7 min.). 

4. 3.868 in. 9. 10.80 ft. 

5. 5.21 ft., 8.24 ft. 10. 12.5 in. 

Bearing Design — Page 158 

1. 571 lbs. per sq. in. 6. 5.006 in. 

2. 6.349 in. 7. 157.5 sq. in. 

3. 20.089 in-. 13-393 in. 8. 7.209 in. 

4. 3.438 in. 9. 3.250 in., 2.167 in. 

5. 10.02 in. 10. 7.5 in., 5.0 in. 

Ball Bearing Design — Page 160 

1. 130 lbs. 5. 0.223 in., 0.158 in. 

2. 13 balls, 26 balls. 6. 12 balls. 

3. 0.130 in. 7. 1. 00 in. 

4. 0.306 in. 8. I5balls. 



ANSWERS TO EXERCISES 



271 



Center of Gravity, Moment of Inertia, etc.^-Page 165 

1. x = 2.167", y — 1. 167". 4. 10.750. 



2. 


x = 1.926", 


y = 1. 148". 5. 4.302. 


3. 


* = 4-937". 


y = 6.214". 

Strength of Materials — Page 186 


I. 


2.236 in. 


6. 36.462 lbs. 


2. 


0.0163 in. 


7. 2.48 in. 


3- 


5 5, ^oo lbs. 


8. 0.020 in. 


4- 


0.249 in. 


9. 2.82 in. 


5- 


3-799 in. 


10. 30,000,000. 
Springs — Page 191 


1. 


3.072 in. 


6. 0.694 in. 


2. 


490.625 lbs. 


7. O.D. = 2 15/32 in 


3- 


3.297 in. 


h = 4.853 in. 


4- 


0.406 in. 


8. 40.502 in. 


5- 


f in. 


9. W = 6,125 lbs., 5 
10. 892.43 lbs. 

Pipes and Cylinders — Page 193 


1. 


0.36 in. 


4. 0.3 in. 


2. 


125 lbs. 


5. 1,318.75 lbs. 



= 1.6 in. 



Riveted Joints — Page 196 



1. d - !" (or &"). P 


= I&" (for 


4. 0.692. 


f rivet). 




5. d - W or |", £ 


11// 
2. 16 . 




(for if" rivet). 


T // -,!// 

3* I t 34 • 




6. 47*,250 lbs. 




Logarithms- 


—Page 199 


I. 2.6902; T.6902. 




11. 2,073. 


2. 3-SI28; .15128. 




12. 75.2. 


3- 1,920. 




13. 7.663. 


4. 93.9. 




14. 0.3079. 


5. 0.003254. 




15- 57-07. 599-4- 


6. 4-350. 




16. 3.785. 4.045. 


7. 0.03408. 




17. 68.11, 618,000. 


8- 433-4- 




18. 0.007673. 


9- 2,159. 




19. 400.9. 


0. 3.48. 




20. 0.0005986. 



272 INDUSTRIAL MATHEMATICS 







Heat- 


-Page 204 


I. 


(a) 73 2/5 F. 




4. 18.63 lbs. 




(b) 44 3/5° F. 




5. 202 deg. F. 




(<0 5°F. 




6. 7.6 lbs. 


2. 


(a) 17 2/9 C. 




7. 10,340 B.t.u. 




(&) 9 4/9° C. 




8. 1.34 gallons. 




(c) 20° C. 




9. 14.67 h.p. 


3- 


(a) 672 F. (abs.). 




10. 172.5 lbs. 




(b) 420 F. (abs.). 




11. 78.8 lbs. 




(c) 285 C. (abs.)- 










Metal Cutting — Page 209 


I. 


3-5 h.p. 




5. 1,325.36 lbs. 


2. 


1. 17 h.p. 




6. 47 ft. per min. 


3- 


1. 14 h.p. 




7. 30.4 ft. per min 


4- 


0.28 h.p. 




8. t£ in. per rev. 



Force, Work, Energy and Momentum — Page 214 

1. 3,498,134 ft. lbs. 9. 42.9 lbs. per ton. 

2. 3,498,134 ft. lbs. 10. io6f h.p. 

3. 309.4 ft. lbs. 11. 12,000 lbs. 

4. 276,631.9 lbs. 12. 8 h.p. 

5. 165 lbs. 13. 850 lbs. 

6. 35.28 kilowatts. 14. 48,500 lbs. 

7. 127.3 h.p. 15. 27.78 ft. per sec. 

8. 30.8 h.p. 16. 850 lbs., 14. 1 h.p. 

Pendulum — Page 230 



I. 


39.1 in. 


6. 


1. 061 sec. 


2. 


32.1909 ft. per sec. per sec. 


7- 


0.1026 in. 


3- 


0.96 sec. 


8. 


1.023 sec. 


4- 


18.04. 


9- 


2.048 rev. per min. 


5- 


1.226 sec. 


10. 


k = 0.75 sec, r = 0.361 
h = 0.458 ft., a = 38|°. 



Review Exercises — Page 236 

1. 3,819.7 rev. per min. 6. 141, 372 lbs. 

2. 6,021.4 ft. per min. 7. 1,380.6 lbs. 

3. 125. 8. 31,406.8 lbs. 
4- 125. 9- 3.305 in. 

5. 0.000355 in. 10. 25. 




Internal Grinding Machine 



THE LITERATURE OF THE 

SCIENCES AND ENGINEERING 



On our shelves is the most complete stock of 
technical, industrial, engineering and scientific 
books in the United States. The technical liter- 
ature of every trade is well represented, as is also 
the literature relating to the various sciences, 
both the books useful for reference as well as 
those fitted for students' use as textbooks. 
A large number of these we publish and for an 
ever increasing number we are the sole agents. 

ALL INQUIRIES MADE OF US ARE CHEER- 
FULLY AND CAREFULLY ANSWERED 
AND COMPLETE CATALOGS AS WELL AS 
SPECIAL LISTS SENT FREE ON REQUEST 




D. VAN NOSTRAND COMPANY 
Publishers and Booksellers 

8 WARREN STREET NEW YORK 



